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THE  LIBRARY 

OF 

THE  UNIVERSITY 
OF  CALIFORNIA 

LOS  ANGELES 


The 


:ALP 

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4 

H  D.  REED  LIBRARY 

TMK.-NT  OP'  OEOLOGY 
ITY  OF  CALIFORNIA. 
ATVGKI,KS,   CALIF. 

Stephen  Rook 

1 

1. 

2. 

3. 

4. 

5. 

6. 

7. 

8. 

9. 

10. 

11. 

12. 

13. 

14. 

15. 

16. 

17. 

18. 

19. 

20. 

21. 

22. 

23. 

24. 

25. 

26. 

27. 

28. 

29. 

30. 

31. 

32. 

33. 

34. 

35. 

36. 

37. 

38. 

39. 

40. 


41.  Thru  Exer.   27  -  pp.   213 

42.  Finish  Chap.   10 
^.4.3  •   Thru  page  219 

/*•«•.  To  art.  4,   Pace  226 

45.  Thru  page  231 

46.  Finish  Chap.  11 
Review  examination. 


47 


Thru  r>p.l5 
"        "     24- 
»       ••     29 
••        "     35      «* 

To  art.  8  pp.41 
Thru  pp.48 
Finish  Chap.  II. 
Thru  pp.59 
Finish  Chap.  3. 
Review  Ex.  Chap.  1-3 
To  art.  3  pp.69 
Thru  np.74 
"   "  77 
To  art.  2  pp.83 
Finish  Chap.  5. 
Graph  .Algebraic  Functions 
"       "       " 
"  Transcendental  " 
"       "       " 

II  Mil  II 

Review  examination 

To  art.  3  pp.93 

Chap.  9  art.  1  and  2 

Art.  3  -  pp«93  thru  examp.10  pp.98 

Finish  Chap.  6 

To  art.  4  -  page  108 

To  art.  7  -  page  113  and  art.  1-chap.ll. 

Thru  pa^e  121 

Finish  Chap.  7. 

Thru  art.  5.  page  133 

•Thru  art.  9  "    142 

Thru  Sxer.l7,-pp.l50 

Finish  Chap.  8. 

Review  examination 

Thru  pp.  171 

Thru  pp.  174 

Chap  .10  r^Thru  pp.197 

Thru  pp.202 

Thru  Bxer.   11  -  pp.209 

Thru     "         10  -  pp.211 


PLANE  AND   SOLID 
ANALYTIC   GEOMETRY 


THE  MACMILLAN  COMPANY 

NEW  YORK    •    BOSTON   •    CHICAGO   •    DALLAS 
ATLANTA   •    SAN    FRANCISCO 

MACMILLAN  &  CO.,  LIMITED 

LONDON    •    BOMBAY   •    CALCUTTA 
MELBOURNE 

THE  MACMILLAN  CO.  OF  CANADA,  LTD. 

TORONTO 


PLANE  AND  SOLID 
ANALYTIC  GEOMETRY 


BY 


WILLIAM  F.  OSGOOD,  PH.D.,  LL.D. 

PERKINS    PROFESSOR    OF    MATHEMATICS 
IN    HARVARD    UNIVERSITY 


AND 

WILLIAM  C.  GRAUSTEIN,  PH.D. 

ASSISTANT    PROFESSOR    OF    MATHEMATICS 
IN    HARVARD    UNIVERSITY 


THE   MACMILLAN   COMPANY 

1921 

All  rights  reserved 


COPYRIGHT,  1920,  1921, 
BY  THE   MACMILLAN  COMPANY. 


Set  up  and  electrotyped.     Published  July,  1921. 


J.  S.  Gushing  Co.  —  Berwick  &  Smith  Co. 
Norwood,  Mass.,  U.S.A. 


Library. 


PREFACE 

THE  object  of  an  elementary  college  course  in  Analytic  Geom- 
etry is  twofold :  it  is  to  acquaint  the  student  with  new  and 
interesting  and  important  geometrical  material,  and  to  provide 
him  with  powerful  tools  for  the  study,  not  only  of  geometry 
and  pure  mathematics,  but  in  no  less  measure  of  physics  in  the 
broadest  sense  of  the  term,  including  engineering. 

To  attain  this  object,  the  geometrical  material-  should  be 
presented  in  the  simplest  and  most  concrete  form,  with  emphasis 
on  the  geometrical  content,  and  illustrated,  whenever  possible, 
by  its  relation  to  physics.  This  principle  has  been  observed 
throughout  the  book.  Thus,  in  treating  the  ellipse,  the  methods 
actually  used  in  the  drafting  room  for  drawing  an  ellipse  from 
the  data  commonly  met  in  descriptive  geometry  are  given  a 
leading  place.  The  theorem  that  the  tangent  makes  equal 
angles  with  the  focal  radii  is  proved  mechanically  :  a  rope  which 
passes  through  a  pulley  has  its  ends  tied  at  the  foci  and  is  drawn 
taut  by  a  line  fastened  to  the  pulley.  Moreover,  the  meaning 
of  foci  in  optics  and  acoustics  is  clearly  set  forth.  Again,  there 
is  a  chapter  on  the  deformations  of  an  elastic  plane  under  stress, 
with  indications  as  to  the  three-dimensional  case  (pure  strain, 
etc.). 

The  methods  of  analytic  geometry,  even  in  their  simplest 
forms,  make  severe  demands  on  the  student's  ability  to  compre- 
hend the  reasoning  of  higher  mathematics.  Consequently,  in 
presenting  them  for  the  first  time,  purely  algebraic  difficulties, 
such  as  are  caused  by  literal  coefficients  and  long  formal  compu- 
tations, should  be  avoided.  The  authors  have  followed  this 
principle  consistently,  beginning  each  new  subject  of  the  early 
chapters  with  the  discussion  of  a  simple  special,  but  typical, 
case,  and  giving  immediately  at  the  close  of  the  paragraph 

788&51 


VI  PREFACE 

simple  examples  of  the  same  sort.  They  have  not,  however, 
stopped  here,  but  through  carefully  graded  problems,  both  of 
geometric  and  of  analytic  character,  have  led  the  student  to 
the  more  difficult  applications  of  the  methods,  and  collections 
of  examples  at  the  close  of  the  chapters  contain  such  as  put 
to  the  test  the  initiative  and  originality  of  the  best  students. 

As  a  result  of  this  plan  the  presentation  is  extraordinarily 
elastic.  It  is  possible  to  make  the  treatment  of  any  given  topic 
brief  without  rendering  the  treatment  of  later  topics  unin- 
telligible, and  thus  the  instructor  can  work  out  a  course  of  any 
desired  extent.  For  example,  one  freshman  course  at  Harvard 
devotes  about  thirty  periods  to  analytic  geometry  and  the  ma- 
terial covered  consists  of  the  essential  parts  of  the  first  nine 
chapters.  Another  freshman  course  gives  twice  the  time  to 
analytic  geometry  (the  students  having  already  had  trigo- 
nometry), taking  up  determinants  and  the  descriptive  properties 
of  the  quadric  surfaces,  and  also  devoting  more  time  to  the  less 
elementary  applications  of  the  methods  of  analytic  geometry/ 
The  advanced  courses  in  the  calculus  and  mechanics  require  the 
material  of  the  later  chapters.  In  fact,  a  thorough  elementary 
treatment  of  the  rudiments  of  Solid  Analytic  Geometry  is  in- 
dispensable for  the  understanding  of  standard  texts  on  applied 
mathematics.  It  is  true  that  these  texts  are  chiefly  Continental. 
But  we  shall  never  have  American  treatises  which  are  up  to  the 
best  scientific  standards  of  the  day  until  the  subjects  above  men- 
tioned are  available  in  simply  intelligible  form  for  the  under- 
graduate. 

The  subject  of  loci  is  brought  in  early  through  a  brief  intro- 
ductory chapter,  and  problems  in  loci  are  spread  throughout 
the  book.  A  later  chapter  is  devoted  to  a  careful  explanation 
of  the  method  of  auxiliary  variables.  There  is  a  chapter  on 
determinants,  with  applications  both  to  analytic  geometry  and 
to  linear  equations.  Diameters  and  poles  and  polars  in  the 
plane  and  in  space  receive  a  thorough  treatment.  Cylindrical 
and  spherical  coordinates  and  quadric  surfaces  are  illumined  by 
the  concept  of  triply  orthogonal  systems  of  surfaces.  The  re- 


PREFACE  Vll 

duction  of  the  general  equation  of  the  second  degree  in  space 
to  normal  forms  by  translations  and  rotations  is  sketched  and 
illustrated  by  numerical  examples. 

The  question  may  be  asked :  In  so  extensive  a  treatment  of 
analytic  geometry  should  not,  for  example,  homogeneous  co- 
ordinates find  a  place  ?  The  authors  believe  that  the  student, 
before  proceeding  to  the  elaborate  methods  of  modern  geometry, 
should  have  a  thorough  knowledge  both  of  the  material  and  the 
methods  which  may  fairly  be  called  elementary,  and  they  felt 
that  a  book  which,  avoiding  the  conciseness  of  some  of  the 
current  texts  and  the  looseness  of  others,  is  clear  because  it  is 
rigorous  will  meet  a  real  need. 

This  book  is  designed  to  be  at  once  an  introduction  to  the 
subject  and  a  handbook  of  the  elements.  May  it  serve  alike 
the  needs  of  the  future  specialist  in  geometry,  the  analyst,  the 
mathematical  physicist,  and  the  engineer.  • 

HARVARD  UNIVERSITY 
.  April,  1921 


CONTENTS 

PLANE  ANALYTIC  GEOMETRY 

INTRODUCTION 
DIRECTED   LINE-SEGMENTS.     PROJECTIONS 

ART.  PAGE 

1.  Directed  Line-Segments 2 

2.  Algebraic  Representation  of  Directed  Line-Segments    .         .  3 

3.  Projection  of  a  Broken  Line 5 

CHAPTER   I 
COORDINATES.     CURVES   AND   EQUATIONS 

1.  Definition  of  Rectangular  Coordinates 7 

2.  Projections  of  a  Directed  Line-Segment  on  the  Axes     .         .  9 

3.  Distance  between  Two  Points 10 

4.  Slope  of  a  Line      .         .     '  •  .         .         .         .         .         .         .12 

5.  Mid-Point  of  a  Line-Segment 16 

0.  Division  of  a  Line-Segment  in  Any  Ratio      ....  17 

7.  Curve  Plotting.     Equation  of  a  Curve          ....  18 

8.  Points  of  Intersection  of  Two  Curves 22 

Exercises  on  Chapter  I        ,._.t    „.<•.,,-,<,,.,{   f, -.    .*•  \,  <\  , -..  •  24 

CHAPTER   II 
THE   STRAIGHT   LINE 

1.  Equation  of  Line  through  Two  Points  .      ''';      '  '.'  '  -'•".' :'     .  27 

2.  One  Point  and  the  Slope  Given     ......  30 

3.  The  General  Equation  of  the  First  Degree    ....  31 

4.  Intercepts      ....      ,,..,„        ....  33 

5.  The  Intercept  Form  of  the  Equation  of  a  Line      ...  35 

6.  Parallel  and  Perpendicular  Lines  .         .         .         .     ,.*.•;;,.  36 

7.  Angle  between  Two  Lines     .         .        .•.>,.<.  M  •<,,.»..-:;••.  38 

8.  Distance  of  a  Point  from  a  Line    .         .  41 


X  CONTENTS 

ART.  PAGE 

9.     Area  of  a  Triangle 43 

10.  General  Theory  of  Parallels  and  Perpendiculars.     Identical 

Lines 44 

11.  Second  Method  of  Finding  Parallels  and  Perpendiculars       .  47 
Exercises  on  Chapter  II        .        .     ' 49 

CHAPTER   III 
APPLICATIONS 

1.  Certain  General  Methods 53 

2.  The  Medians  of  a  Triangle 54 

3.  Continuation.     The  General  Case         .'  .        .        .56 

4.  The  Altitudes  of  a  Triangle 58 

5.  The  Perpendicular  Bisectors  of  the  Sides  of  a  Triangle          .  60 

6.  Three  Points  on  a  Line         .         .        . . .      .        .        .        .60 

Exercises  on  Chapter  III 61 

CHAPTER   IV 
THE   CIRCLE 

1.  Equation  of  the  Circle 65 

2.  A  Second  Form  of  the  Equation 66 

3.  Tangents 69 

4.  Circle  through  Three  Points          .         .         .         .         .         .72 

Exercises  on  Chapter  IV 74 

CHAPTER  V       . 

INTRODUCTORY  PROBLEMS   IN   LOCI.      SYMMETRY   OF 
CURVES 

1.  Locus  Problems 79 

2.  Symmetry 83 

Exercises  on  Chapter  V 86 

CHAPTER  VI 
THE   PARABOLA 

1.  Definition 88 

2.  Equation  of  the  Parabola 90 

3.  Tangents 93 


CONTENTS  XI 

ART.  PAGE 

4.     Optical  Property  of  the  Parabola 95 

Exercises  on  Chapter  VI 97 

CHAPTER  VII 
THE   ELLIPSE 

1.  Definition 

2.  Geometrical  Construction 

3.  Equation  of  the  Ellipse          .         .         .         . 

4.  Tangents 

5.  Optical  and  Acoustical  Meaning  of  the  Foci          , 

6.  Slope  and  Equation  of  the  Tangent 

7.  A  New  Locus  Problem  ....... 

8.  Discussion  of  the  Case  e<l.     The  Directrices 

9.  The  Parabola  as  the  Limit  of  Ellipses 

10.  New  Geometrical  Construction  for  the  Ellipse.     Parametric 

Representation 119 

Exercises  on  Chapter  VII 120 

CHAPTER   VIII 
THE   HYPERBOLA 

1.  Definition 124 

2.  Equation  of  the  Hyperbola 125 

3.  Axes,  Eccentricity,  Focal  Radii 127 

4.  The  Asymptotes 129 

5.  Tangents .         .         .133 

6.  New  Definition.     The  Directrices 136 

7.  The  Parabola  as  the  Limit  of  Hyperbolas.     Summary .         .  139 

8.  Hyperbolas  with  Foci  on  the  Axis  of  y.   Conjugate  Hyperbolas  141 

9.  Parametric  Representation 142 

10.  Conic  Sections       .         .         .'        ...         .        .         .     144 

11.  Confocal  Conies 145 

Exercises  on  Chapter  VIII ;     ,  ..      .     148 


1.  Tangents       .        .        .        .        .        .•  ^ •  .  -•:"'..   ~.        .     154 

2.  Continuation.     Implicit  Equations       .        .-  '.        .        .     160 


Xil  CONTENTS 

ART.  PAGE 

3.  The  Equation  u  +  kv  =  0 165 

4.  The  Equation  vv  =  0 .173 

5.  Tangents  with  a  Given  Slope.     Discriminant  of  a  Quadratic 

Equation 174 

6.  General  Formulas  for  Tangents  with  a  Given  Slope      .         .  179 

7.  Tangents  to  a  Conic  from  an  External  Point         .         .         .  185 
Exercises  on  Chapter  IX 188 

CHAPTER  X 
POLAR   COORDINATES 

1.  Definition      ..........  193 

2.  Circles 194 

3.  Straight  Lines       .         .        . 196 

4.  Graphs  of  Equations     .         .      ' 198 

5.  Conies .    .         .202 

6.  Transformation  to  and  from  Cartesian  Coordinates      .         .  206 
Exercises  on  Chapter  X 210 

CHAPTER  XI 
TRANSFORMATION   OF   COORDINATES 

1.  Parallel  Axes 216 

2.  Rotation  of  the  Axes 219 

3.  The  General  Case 223 

4.  Determination  of  the  Transformation  from  the  Equations  of 

the  New  Axes k  .      .  226 

5.  Reversal  of  One  Axis 230 

Exercises  on  Chapter  XI 231 

CHAPTER  XII 
THE   GENERAL  EQUATION   OF  THE   SECOND   DEGREE 

1.  Change  of  Origin  of  Coordinates 235 

2.  Rotation  of  Axes 239 

3.  Continuation.     General  Case 241 

4.  The  General  Equation,  &  -  4  AC  =£  0 247 

5.  The  General  Equation,  &-<!  AC  =  0 250 

6.  Summar)'.     Invariants 256 

Exercises  on  Chapter  XII     .        .        .        .        .        .        .  257 


CONTENTS  Xlll 

CHAPTER  XIII 

A   SECOND    CHAPTER   ON   LOCI.     AUXILIARY   VARIABLES. 
INEQUALITIES 

ART.  PAGE 

1.  Extension  of  the  Method  for  the  Determination  of  Loci        .  261 

2.  One  Auxiliary  Variable 263 

3.  Coordinates   of   a   Point   Tracing   a   Curve,    as   Auxiliary 

Variables 266 

4.  Other  Problems  Involving  Two  or  More  Auxiliary  Variables  272 

5.  Use  of  the  Formula  for  the  Sum  of  the  Roots  of  a  Quadratic 

Equation 274 

6.  Loci  of  Inequalities       ........  277 

7.  Locus  of  Two  or  More  Simultaneous  Inequalities          .         .  279 

8.  Bisectors  of  the  Angles  between  Two  Lines  .        .        .        .281 
Exercises  on  Chapter  XIII    .         .         .        .        .        .         .283 

CHAPTER  XIV 
DIAMETERS.     POLES   AND   POLARS 

1.  Diameters  of  an  Ellipse     ' 288 

2.  Conjugate  Diameters  of  an  Ellipse 290 

3.  Diameters  of  a  Hyperbola 293 

4.  Conjugate  Diameters  of  a  Hyperbola 295 

5.  Diameters  of  a  Parabola 299 

6.  Extremities  and  Lengths  of  Conjugate  Diameters        .^       .  299 

7.  Physical  Meaning  of  Conjugate  Diameters   .         .        .        .  302 

8.  Harmonic  Division        .         .        .,••,  '.    ,    .     ,   ,.   .     .         .  309 

9.  Polar  of  a  Point    .        .        .        .'""."      .'      ".''''.''".  311 

10.  Pole  of  a  Line 315 

11.  Properties  of  Poles  and  Polars       .       •_. ,    •/»,  i  ',•        •        •  317 

12.  Relative  Positions  of  Pole  and  Polar     .        V'"'.        .         .320 

13.  Construction  Problems          -j ' -.  i  -'».        .         .         .         .         .  324 

Exercises  on  Chapter  XIV    .......  326 

CHAPTER  XV 
TRANSFORMATIONS   OF   THE  PLANE.     STRAIN 

1.  Translations.       ..        .-.'  :.:  •     .  •   ..i. '<••     •  •[  /,»  •  .....:     .  330 

2.  Rotations .•-,,-„,*•:.  332 

3.  Transformations  of  Similitude       .,..:..,.-••      ;!•»         •  334 

4.  Reflections  in  the  Axes                                                     .        .  336 


XIV  CONTENTS 

ABT.  PAGE 

5.  Simple  Elongations  and  Compressions 337 

6.  The  General  Affine  Transformation 342 

7.  Factorization  of  Particular  Transformations         .        .         .  349 

8.  Simple  Shears 351 

9.  Second   Method   of   Factorization.     Homogeneous   Strains  355 
Exercises  on  Chapter  XV 358 

CHAPTER  XVI 
DETERMINANTS   AND   THEIR  APPLICATIONS 

I.  DETERMINANTS 

1.  Simultaneous  Linear  Equations 360 

2.  Two-  and  Three-Rowed  Determinants 361 

3.  Determinants  of  the  Fourth  and  Higher  Orders    .         .         .  364 

4.  Evaluation  of  a  Determinant  by  Minors       ....  367 

5.  Simplified  Evaluation  by  Minors 370 

6.  Fundamental  Properties  of  Determinants     .         .         .         .  373 

7.  Interchanges  of  Rows  and  of  Columns          ....  376 

8.  Cramer's  Rule 381 

9.  Three  Equations  in  Two  Unknowns.     Compatibility   .         .  383 

10.  Homogeneous  Linear  Equations 387 

II.  APPLICATIONS 

11.  The  Straight  Line 391 

12.  The  Circle  and  the  Conies     ....      -^-^.         .  394 
Exercises  on  Chapter  XVI '     .  399 


SOLID  ANALYTIC  GEOMETRY 

CHAPTER  XVII 
PROJECTIONS.     COORDINATES 

1.  Directed  Line-Segments 405 

2.  Projection  of  a  Broken  Line  .......  405 

3.  The  Angle  between  Two  Directed  Lines        ....  406 

4.  Value  of  the  Projection  of  a  Directed  Line-Segment     .         .  408 

5.  Coordinates 409 

6.  Projections  of  a  Directed  Line-Segment  on  the  Axes     .         .  413 

7.  Distance  between  Two  Points  ....  414 


CONTENTS  XV 

ART.  PAGE 

8.  Mid-Point  of  a  Line-Segment        . '       .         .        .  '      .        .  415 

9.  Division  of  a  Line-Segment  in  a  Given  Ratio        .         .         .  416 
Exercises  on  Chapter  XVII .  417 

CHAPTER   XVIII 
DIRECTION   COSINES.     DIRECTION   COMPONENTS 

1.  Direction  Cosines  of  a  Directed  Line 420 

2.  Angle  between  Two  Directed  Lines 425 

3.  Direction  Components  of  an  Undirected  Line        .         .         .  427 

4.  Formulas  for  the  Use  of  Direction  Components    .         .         .  433 

5.  Line  Perpendicular  to  Two  Given  Lines        ....  436 

6.  Three  Lines  Parallel  to  a  Plane 440 

Exercises  on  Chapter  XVIII 441 

CHAPTER  XIX 
THE   PLANE 

1.  Surfaces  and  Equations 444 

2.  Plane  through  a  Point  with  Given  Direction  of  its  Normals  447 

3.  The  General  Equation  of  the  First  Degree    ....  448 

4.  Intercepts 450 

5.  Intercept  Form  of  the  Equation  of  a  Plane  ....  451 

6.  Plane  through  Three  Points 453 

7.  Perpendicular,  Parallel,  and  Identical  Planes.    Angle  between 

Two  Planes 454 

8.  Planes  Parallel  or  Perpendicular  to  a  Given  Plane         .         .  456 

9.  Distance  of  a  Point  from  a  Plane 460 

10.     Point  of  Intersection  of  Three  Planes 462 

Exercises  on  Chapter  XIX    ...'....  465 

CHAPTER  XX 
THE    STRAIGHT   LINE 

1.  Equations  of  a  Curve <  470 

2.  Line  of  Intersection  of  Two  Planes        .         .         .         .         .  476 

3.  Line  through  a  Point  with  Given  Direction  Components      .  478 

4.  Line  through  Two  Points      .......  481 

5.  Line  or  Plane  in  Given  Relationship  to  Given  Lines  or 

Planes  483 


xvi  CONTENTS 

ART.  PAGE 

6.  Angle  between  a  Line  and  a  Plane 487 

7.  Point  of  Intersection  of  a  Line  and  a  Plane  ....  488 

8.  Parametric  Representation  of  a  Curve 490 

Exercises  on  Chapter  XX     .        .        .        .        .        .        .  494 

CHAPTER  XXI 

THE   PLANE   AND   THE   STRAIGHT   LINE.    ADVANCED 
METHODS 

1.  Linear  Combination  of  Two  Planes 498 

2.  Three  Planes  through  a  Line.     Three  Points  on  a  Line         .  504 

3.  Line  in  a  Plane      .........  506 

4.  Four  Points  in  a  Plane.     Four  Planes  through  a  Point         .  507 

5.  Two  Intersecting  Lines 512 

6.  Distance  of  a  Point  from  a  Line.     Distance  between  Two 

Lines 514 

7.  Area  of  a  Triangle.     Volume  of  a  Tetrahedron     .         .         .  516 
Exercises  on  Chapter  XXI 519 

CHAPTER  XXII 

SPHERES,   CYLINDERS,   CONES.     SURFACES   OF 
REVOLUTION 

1.  Equation  of  the  Sphere 523 

2.  General  Form  of  the  Equation 524 

3.  Sphere  through  Four  Points 526 

4.  Tangent  Plane  to  a  Sphere 527 

5.  The  Circle     ...... 529 

6.  Cylinders 532 

7.  Cones    . 536 

8.  Surfaces  of  Revolution 540 

Exercises  on  Chapter  XXII 544 

CHAPTER  XXIII 
QUADRIC   SURFACES 

1.  The  Ellipsoid 548 

2.  The  Hyperboloids 550 

3.  The  Paraboloids 553 

4.  Rulings 555 


CONTENTS  XVll 

ART.  PAGE 

5.  Parallel  Sections 561 

6.  Circular  Sections 564 

7.  Tangent  Lines  and  Planes .  566 

8.  Diameters.     Diametral  Planes 569 

9.  Poles  and  Polars 575 

10.     One-Dimensional  Strains,  with  Applications          .         .         .  577 

Exercises  on  Chapter  XXIII 580 

CHAPTER  XXIV 

SPHERICAL   AND    CYLINDRICAL   COORDINATES. 
TRANSFORMATION    OF    COORDINATES 

1.  Spherical  Coordinates   ........  584 

2.  Cylindrical  Coordinates 587 

3.  Triply  Orthogonal  Systems  of  Surfaces         ....  589 

4.  Confocal  Quadrics 590 

5.  Transformation  to  Parallel  Axes 592 

6.  Rotation  of  the  Axes 593 

7.  The  General  Equation  of  the  Second  Degree         .        .         .  599 
Exercises  on  Chapter  XXIV 604 


PLANE  ANALYTIC  GEOMETRY 

» 

INTRODUCTION 

DIRECTED  LINE-SEGMENTS.    PROJECTIONS 

Elementary  Geometry,  as  it  is  studied  in  the  high  school 
to-day,  had  attained  its  present  development  at  the  time  when 
Greek  culture  was  at  its  height.  The  first  systematic  treat- 
ment of  the  subject  which  has  come  down  to  us  was  written 
by  Euclid  about  300  B.C. 

Algebra,  on  the  other  hand,  was  unknown  to  the  Greeks. 
Its  beginnings  are  found  among  the  Hindus,  to  whom  the  so- 
called  Arabic  system  of  numerals  may  also  be  due.  It  came 
into  Western  Europe  late,  and  not  till  the  close  of  the  middle 
ages  was  it  carried  to  the  point  which  is  marked  by  any  school 
book  of  to-day  that  treats  this  subject. 

When  scholars  had  once  possessed  themselves  of  these  two 
subjects  —  Geometry  and  Algebra  —  the  next  step  was  quickly 
taken.  The  renowned  philosopher  and  mathematician,  Rene 
Descartes,  in  his  Geometric  of  1637,  showed  how  the  methods 
of  algebra  could  be  applied  to  the  study  of  geometry.  He 
thus  became  the  founder  of  Analytic  Geometry.* 

The  "  originals "  and  the  locus  problems  of  Elementary 
Geometry  depend  for  their  solution  almost  wholly  on  ingenu- 
ity. There  are  no  general  methods  whereby  one  can  be  sure 
of  solving  a  new  problem  of  this  class.  Analytic  Geometry, 

*Also  called  Cartesian  Geometry,  from  the  Latinized  form  of  his  name, 
Cartesius. 

1 


2  ANALYTIC  GEOMETRY 

on  the  other  hand,  furnishes  universal  methods  for  the  treat- 
ment of  such  problems ;  moreover,  these  methods  make  pos- 
sible the  study  of  further  problems  not  thought  of  by  the 
ancients,  but  lying  at  the  heart  of  modern  mathematics  and 
mathematical  physics.  Indeed,  these  two  great  subjects  owe 
their  very  existence  to  the  new  geometry  and  the  Calculus. 

The  question  of  how  to  make  use  in  geometry  of  the  nega- 
tive, as  well  as  the  positive,  numbers  is  among  the  first  which 
must  be  answered  in  applying  algebra  to  geometry.  The  solu- 
tion of  this  problem  will  become  clear  in  the  following 
paragraphs. 

1.   Directed  Line-Segments.     Let  an  indefinite  straight  line, 

L,  be  given,  and  let  two  points,  A  and  B,  be  marked  on  L. 

.  -D          f,       Then  the  portion  of  L  which  is 

1 1 1 bounded  by  A  and  B  is  what  is 

A          Q     g  called  in  Plane  Geometry  a  line- 

' ' ' segment,  and  is  written  as  AB. 

Q      A  B  -^e^ a  third  point,  C,  be  marked 

'         ' ' on  L.     Then  three  cases  arise, 

as  indicated  in  the  figure.     Cor- 
responding to  these  three  cases  we  have  : 

(a)  AB  +  BC  =  AC; 

(6)  AB-CB  =  AC; 

(c)  CB-AB=  CA. 

Three  other  cases  will  arise  if  the  original  points  A  and  B  are 
taken  in  the  opposite  order  on.  the  line.  Let  the  student 
write  down  the  three  corresponding  equations. 

A  unification  of  all  these  cases  can  be  effected  by  means  of 
an  extension  of  the  concept  of  a  line-segment.  We  no  longer 
consider  the  line-segments  AB  and  BA  as  identical,  but  we 
distinguish  between  them  by  giving  each  a  direction  or  sense. 
Thus,  AB  shall  be  directed  from  A  to  B  and  BA  shall  be 
directed  from  B  to  A,  i.e.  oppositely  to  AB.  These  directed 


INTRODUCTION  3 

line-segments  we  denote  by  AB  and  BA,  to  distinguish  them 
from  the  ordinary,  or  undirected,  line-segments. 

We  may,  for  the  moment,  interpret  the  directed  line-seg- 
ment AB  as  the  act  of  walking  from  A  to  B ;  then  BA  repre- 
sents the  act  of  walking  from  B  to  A.  With  this  in  mind,  let 
us  return  to  Fig.  1  and  consider  the  directed  line-segments 
AB,  BC,  and  AC.  We  have,  in  all  three  cases  represented  by 
Fig.  1,  and  also  in  the  other  three : 

AB  +  BC=AC, 

since  walking  from  A  to  B  and  then  walking  from  B  to  C  is 
equivalent,  with  reference  to  the  point  reached,  to  walking 
from  A  to  C. 

Accordingly,  we  unify  all  six  cases  by  defining,  as  the  sum 
of  the  directed  line-segments  AB  and  BC,  the  directed  line- 
segment  AC : 

(1)  AB  +  BC=AC. 

From  this  definition  it  follows  that,  if  A,  B,  C,  and  D  are 
any  four  points  of  L, 

(2)  AB  +  BC+CD  =  AD. 

For,  by  (1),  the  sum  of  the  first  two  terms  in  (2)  is  AC,  and, 
by  the  definition,  the  sum  of  AC  and  CD  is  AD. 

Similarly,  if  the  points  M,  M1}  M2,  •  •  •,  Mn_»  N-  are  any  points 
of  L,  we  have 


(3)        MMl  +  MiMz  +  •  •  •  +  JGH*+  +  Mn_^N  =  MN. 

Given  two  directed  line-segments  on  the  same  line  or  on 
two  parallel  lines,  we  say  that  these  two  directed  line-segments 
are  equal,  if  they  have  equal  lengths  and  the  same  direction  or 

sense. 

2.   Algebraic  Representation  of  Directed  Line-Segments.     On 

the  line  L  let  one  of  the  two  opposite  directions  or  senses  be 
chosen  arbitrarily  and  defined  as  the  positive  direction  or  sense 
of  L ;  and  let  the  other  be  called  the  negative  direction  or  sense. 


4  ANALYTIC  GEOMETRY 

A  directed  line-segment  AB,  which  lies  on  L,  is  then  called 
positive,  if  its  sense  is  the  same  as  the  positive  sense  of  L,  and 
negative,  if  its  sense  is  the  same  as  the  negative  sense  of  L. 

To  such  a  directed  line-segment  AB  we  assign  a  number, 
which  we  shall  also  represent  by  AB,  as  follows.  If  I  is  the 
length  of  the  ordinary  line-segment  AB,  then 

AB  =  I,        if  AB  is  a  positive  line-segment  ; 
AB  =  —  l,     if  AB  is  a  negative  line-segment. 

If  AB  =  I,  then  BA  =  -  I  ;  and  if  AB  =  —  I,  then  ~BA=  I 
In  either  case 


(1)  AB  +  BA  =  Q         or        AB  =  -BA. 

Since  the  act  of  walking  from  A  to  B  is  nullified  by  the  act 
of  walking  from  B  to  A,  we  might  have  arrived  at  equations 
(1)  from  consideration  of  the  line-segments  themselves,  instead 
of  by  use  of  the  numbers  which  represent  them. 

It  is  easy  to  verify  the  fact  that  equations  (1),  (2),  and  (3) 
of  the  preceding  paragraph,  which  relate  to  directed  line-seg- 
ments, hold  for  the  corresponding  numbers.  Consequently, 
no  error  or  confusion  arises  from  using  the  same  notation  AB 
for  both  the  directed  line-segment  and  the  number  correspond- 
ing to  it.  We  shall,  however,  adopt  a  still  simpler  notation, 
dropping  the  dash  altogether  and  writing  henceforth  AB  to 
denote,  not  merely  the  directed  line-segment  or  the  number 
corresponding  to  it,  but  also  the  line-segment  itself,  stating 
explicitly  what  is  meant,  unless  the  meaning  is  clear  from  the 
context. 

Absolute  Value.  It  is  often  convenient  to  be  able  to  express 
merely  the  length  of  a  directed  line-segment,  AB.  The  nota- 
tion for  this  length  is  |  AB  \  ;  read  :  "  the  absolute  value  of 
AB." 

The  numerical,  or  absolute,  value  of  a  number,  a,  is 
denoted  in  the  same  way  :  |  a  |.  Thus,  |  —  3  1  =  3.  Of  course, 
13|=3. 


INTRODUCTION 


3.  Projection  of  a  Broken  Line.  By  the  projection  of  a  point 
P  on  a  line  L  is  meant  the  foot,  M,  of  the  perpendicular 
dropped  from  P  on  L.  If  P  lies  on  L,  it  is  its  own  projection 
on  L. 

Let  PQ  be  any  directed  line-segment,  and  let  L  be  an  arbi- 
trary line.  Let  M  and  N  be  respectively  the  projections  of 
P  and  Q  on  L.  The  projection 
of  the  directed  line-segment  PQ 
on  i  shall  be  denned  as  the  di- 
rected line-segment  MN,  or  the 
number  which  represents  MN  al- 
gebraically. Since  MN=  —  NM, 
it  follows  that 


M 


N 
FIG.  2 


Proj.  PQ  =  -Proj.  QP. 


If  PQ  lies  on  a  line  perpendicular  to  L,  the  points  M  and 
^T  coincide,  and  we  say  that  the  projection  MN  of  PQ  on  L  is 
zero.  Such  a  directed  line-segment  MN,  whose  end-points  are 
identical,  we  may  call  a  nil-segment ;  to  it  corresponds  the 
number  zero.  It  is  evident  that  in  taking  the  sum  of  a  num- 
ber of  directed  line-segments,  any  of  them  which  are  nil- 
segments  may  be  disregarded,  just  as,  in  taking  the  sum  of  a 

set  of  numbers,  any  of  them 
which  are  zero  may  be  disre- 
garded. 

Consider  an  arbitrary 
broken  line  PP^  •  •  •  Pn_iQ. 
By  its  projection  on  L  is 
meant  the  sum  of  the  pro- 
jections of  the  directed  line- 


FIG.  3 


segments  PPX,  PtP2,  •  •  •,  Pn_iQ,  or 


This  sum  has  the  same  value  as  MN,  the  projection  on  L 
of  the  directed  line-segment  PQ  ;  cf.  §  1,  (3)  : 

MMl  +  M^2  +  •  •  •  +  Mn_lN  =  MN. 


6  ANALYTIC  GEOMETRY 

Hence  the  theorem  : 

THEOREM  1.  The  sum  of  the  projections  on  L  of  the  segments 
PPi,  P\PK  •  •  -,  Pn-iQ  of  a  broken  line  joining  P  with  Q  is  equal 
to  the  projection  on  L  of  the  directed  line-segment  PQ. 

If,  secondly,  the  same  points  P  and  Q  be  joined  by  another 
broken  line,  PP[P'2  •  •  •  PL-iQ,  the  projection  of  the  latter  on 
L  will  also  be  equal  to  MN: 

MM[  +  M[M'2+  ».  +  M'm_lN=  MN. 
Hence  the  theorem  : 

THEOREM  2.     Given  two  broken  lines  having  the  same  extremi- 

ties, 

•  P  and 


Let  L  be  an  arbitrary  straight  line.  Then  the  sum  of  the  pro- 
jections on  L  of  the  segments  PP\,  PiPz)  •  •  •>  Pn-iQ}  of  which  the 
first  broken  line  is  made  up,  is  equal  to  the  corresponding  sum 
for  the  second  broken  line. 


CHAPTER   I 


N 


O 


y 


COORDINATES.     CURVES  AND  EQUATIONS 

1.  Definition  of  Rectangular  Coordinates.  Let  a  plane  be 
given,  in  which  it  is  desired  to  consider  points  and  curves. 
Through  a  point  0  in  this  plane  take  two  indefinite  straight 
lines  at  right  angles  to  each  other,  and  choose  on  each  line  a 
positive  sense. 

Let  P  be  any  point  of  the  plane.  Consider  the  directed 
line-segment  OP.  Let  its  projections  on  the  two  directed  lines 
through  0  be  OM  and  ON.  The  numbers  which  represent 
algebraically  these  projections, 
that  is,  the  lengths  of  OM  and 
ON  taken  with  the  proper  signs 
(cf.  Introduction,  §  2),  are  called 
the  coordinates  of  P.  We  shall 
denote  them  by  x  and  y : 

x  =  OM,  y  =  ON, 

and  write  them  in  parentheses : 

(a?,  y).     The  first  number,  x,  is 

known    as    the    x-coo'rdinate,    or 

abscissa,  of  P;  the  second,  y,  as  the  y-coordinate,  or  ordinate, 

of  P. 

The  point  0  is  called  the  origin  of  coordinates.  The  directed 
lines  through  0  are  called  the  axes  of  coordinates  or  the  coordi- 
nate axes  ;  the  one,  the  axis  of  x  ;  the  other,  the  axis  of  y.  It 
is  customary  to  take  the  coordinate  axes  as  in  Fig.  1,  the  axis 
of  x  being  positive  from  left  to  right,  and  the  axis  of  y,  posi- 
tive from  below  upward.  But,  of  course,  the  opposite  sense 
on  one  or  both  axes  may  be  taken  as  positive,  and  an  oblique 


M 


FIG.  1 


Q 


8  ANALYTIC  GEOMETRY 

position  of  the  axes  which  conforms  to  the  definition  is  legiti- 
mate, the  essential  thing  being  solely  that  the  axes  be  taken 
perpendicular  to  each  other. 

Every  point,  P,  in  the  plane  has  definite  coordinates,  (x,  y). 
Conversely,  to  any  pair  of  numbers,  x  and  y,  corresponds  a 
point  P  whose  coordinates  are  (x,  y).  This  point  can  be  con- 
structed by  laying  off  OM=x  on  the  axis  of  x,  erecting  a  per- 
pendicular at  M  to  that  axis,  and  then  laying  off  MP  =  y. 
We  might  equally  well  have  begun  by  laying  off  ON=  y  on 
the  axis  of  y  (cf.  Fig.  1),  and  then  erected  a  perpendicular  to 
that  axis  at  N  and  laid  off  on  it 
NP  =  x.  It  shall  be  understood  that 
the  positive  sense  on  any  line  parallel 
to  one  of  the  coordinate  axes,  such  as 
— x  the  perpendicular  to  the  axis  of  x  at 
M,  shall  be  the  same  as  the  positive 

"PTn     9 

sense  of  that  axis.     For  other  lines 

of  the  plane  there  is-  no  general  principle  governing  the  choice 
of  the  positive  sense. 

The  coordinates  of  the  origin  are  (0,  0).  Every  point  on 
the  axis  of  x  has  0  as  its  ordinate,  and  these  are  the  only 
points  of  the  plane  for  which  this  is  true.  Hence  the  axis  of 
x  is  represented  by  the  equation 

y  =  0,  (axis  of  x). 

Similarly,  the  axis  of  y  is  represented  by  the  equation 
x  =  0,  (axis  of  y). 

The  axes  divide  the  plane  into  four  regions,  called  quadrants. 
The  first  quadrant  is  the  region  included  between  the  positive 
axis  of  x  and  the  positive  axis  of  y ;  the  second  quadrant,  the 
region  between  the  positive  axis  of  y  and  the  negative  axis  of 
x ;  etc.  It  is  clear  that  the  coordinates  of  a  point  in  the  first 
quadrant  are  both  positive ;  that  a  point  of  the  second  quad- 
rant has  its  abscissa  negative  and  its  ordinate  positive ;  etc. 

The  system  of  coordinates  just  described  is  known  as  a  sys- 
tem of  rectangular  or  Cartesian  coordinates. 


COORDINATES.   CURVES  AND  EQUATIONS 


EXERCISES 

The  student  should  provide  himself  with  some  squared 
paper  for  working  these  and  many  of  the  later  exercises  in 
this  book.  Paper  ruled  to  centimeters  and  subdivided  to  mil- 
limeters is  preferable. 

1.  Plot  the  following  points,  taking  1  cm.  as  the  unit : 
(a)  (0,1);  (6)   (1,0);  (c)    (1,1); 
(d)  (1, -1);               (e)   (-!,-!);          (/)  (2, -3); 

(90    (0,  -21)  ;  (ft)  (-  3.7,  0)  ;  (f)    (-lj,  -If) . 

(.;)  (-4,3.2);  (*)  (3.24,  -0.87);     0     (-1,1). 

2.  Determine  the  coordinates  of  the  point  P  in  Fig.  1  when 
1  in.  is  taken  as  the  unit  of  length ;  also  when  1  cm.  is  the 
unit  of  length. 

3.  The   same   for   the    point   marked    by    the    period    in 
«  Fig.  1." 

2.   Projections  of  a  Directed  Line-Segment  on  the  Axes.     Let 

PI,  with  the  coordinates  (xl}  y^),  and  P2 :  (o^,  y2)*  be  any  two 
points  of  the  plane.  Con- 
sider the  directed  line-seg- 
ment PiP2.  It  is  required 
to  find  its  projections  on  the 
axes. 

To  do  this,  draw  the 
broken  line  P^P.,.  By  In- 
troduction, §  3,  Th.  1,  the 
projections  of  this  broken 
line  on  the  axes  are  the 

same  as   those  of   the   directed   line-segment   PiP2. 
taking  first  the  projections  on  the  axis  of  x,  we  have 


0 


FIG.  3 


Hence, 


Proj. 


=      Proj. 
=  -Proj. 


*  We  shall  frequently  use  this  shorter  notation,  P2  : 
breviation  for  "  P2,  with  the  coordinates  (x2,  2/2)-" 


OP2 
OP2. 

(xj,  2/2),  as  an  ab- 


10 ANALYTIC  GEOMETRY 

But  the  terms  in  the  last  expression   are  by  definition  —  x± 
and  a^.     So 

(1)  Proj.  PiP2  on  ovaxis  =  a%  —  x^ 
Similarly, 

(2)  Proj.  PjP2  on  y-axis  =  t/2  —  yx. 

The  projections  of  PiP2  on  two  lines  drawn  parallel  to  the 
axes  are  obviously  given  by  the  same  expressions. 

EXERCISES 

1.  Plot  PjP2  when  Pj  is  the  point  (a)  of  Ex.  1,  §  1,  and  P2 
is  (6).     Determine  the  projections  from  the  foregoing  formulas, 
and  verify  directly  from  the  figure. 

2.  The  same,  when 

i)  PI  is  (e)  and  P2  is  (/)  ; 
ii)  P!  is  (c)  and  P2  is  (d)  ; 
iii)  PI  is  (i)  and  P2  is  (I). 

3.  Distance  between  Two  Points.     Let  the  points  be  P1} 
with   the   coordinates  (a^,  2h),  and  P2 :  (a^,   y2).     Through  Pj 
draw  a  line  parallel  to  the  axis  of  x  and  through  P2,  a  line 

parallel    to   the  axis  of    y\    let   Q 

y  v^v^v     denote  the  point  of  intersection  of 

these   lines.      Then,   by   the   Pytha- 
gorean Theorem, 

-*    (!) 


O 

FIG.  4  or 

(2) 
where  D  denotes  the  distance  between  Pt  and  P2.     Hence 

(3)  D  = 


In  the  foregoing  analysis,  we  have  used  PtQ  (and  similarly, 
QP2)  in  two  senses,  namely,  i)  as  the  length  of  the  ordinary 
line-segment  P^ty  of  Elementary  Geometry ;  ii)  as  the  algebraic 


COORDINATES.   CURVES  AND  EQUATIONS    11 

expression  x2  —  xl  for  the  projection  P^Q,  of  the  directed  line 
segment  P^P2  on  a  parallel  to  the  axis  of  x.  Since,  however, 
these  two  numbers  differ  at  most  in  sign,  their  squares  are 
equal,  and  hence  equation  (2)  is  equivalent  to  equation  (1). 

In  particular,  PtP2  may  be  parallel  to  an  axis,  e.  g.  the  axis 
of  x.     Here,  y2  =  yl}  and  (3)  becomes 


The  student  must  not,  however,  hastily  infer  that 

D  =  X.2  —  05i. 

It  may  be  that  x2  —  xt  is  negative,  and  then  * 
D  =  -(x2-x1). 

A  single  formula  which  covers  both  cases  can  be  written  in 
terms  of  the  absolute  value  (cf .  Introduction,  §  2)  as  follows : 

(4)  D=\x2-x1\. 

EXERCISES 

1.  Find  the  distances  between  the  following  pairs  of  points, 
expressing  the  result  correct  to  three  significant  figures.     Draw 
a  figure  each  time,  showing  the  points  and  the  line  connecting 
them,  and  verify  the  result  by  actual  measurement. 

(a)     (2,  1)  and  (-  2,  -  2).          (6)   (-  7,  6)  and  (2,  -  3). 
(c)    (13,  5)  and  (-  2,  5).  (d)        (7,  3)  and  (12,  3). 

(e)      (4,  8)  and  (4,  -  8).  (/)  (-  1,  2)  and  (-  1,  6). 

2.  Find  the  lengths  of  the  sides  of  the  triangle  whose  ver- 
tices are  the  points  (—  2,  3),  (— 2,  —  1),  (4,  —  1). 

3.  How  far  are  the  vertices  of  the  triangle  in  question  2 
from  the  origin  ? 

*  There  is  no  contradiction  here,  or  conflict  with  the  ordinary  laws  of 
algebra.  For,  the  -^-sign  always  calls  for  the  positive  square  root,  —  that 
being  the  definition  of  the  symbol,  —  and  we  must  see  to  it  in  any  given 
case  that  we  fulfill  the  contract. 


12 


ANALYTIC  GEOMETRY 


4.  Find  the  lengths  of  the  diagonals  of  the  convex  quadri- 
lateral whose  vertices  are  the  points  (4,  1),  (1,  3),  (—3,  1), 
(-2,-!). 

4.  Slope  of  a  Line.  By  the  slope,  A,  of  a  line  is  meant  the 
trigonometric  tangent  of  the  angle,  6,  which  the  line  makes 

with  the  positive  axis  of  x  : 

(1)  A  =  tan  0. 

To  find  the  slope  of  the  line, 
let  P0  with  the  coordinates  (xiy  ?/i), 
and  P2  :  fa,  y2)  be  the  extremities 
of  any  directed  line-segment 
on  the  line.  Then 


0 


(2) 
or 
(3) 


If,  instead  of  P^,  we  had  taken  its  opposite,  P2P!,  we 
should  have  obtained  for  A  the  value  (yt  —  yz)/(%i  —  ty)-  But 
this  is  equal  to  the  value  of  A  given  by  (3).  Thus,  A  is  the 
same,  whether  the  line  is  directed  in  the  one  sense  or  in  the 
opposite  sense.  Hence  we  think  of  A  as  the  slope  of  the  line 
without  regard  to  sense. 

Variation  of  the  Slope.  Consider  the  slopes,  A,  of  different 
lines,  L,  through  a  given  point,  P.  When  L  is  parallel  to  the 
axis  of  x,  A  has  the  value  zero.  When  L  rotates  as  shown  in 
the  figure,  A  becomes  positive  and  increases  steadily  in  value. 
As  L  approaches  the  vertical  line  L',  A  becomes  very  large, 
increasing  without  limit. 

When  L  passes  beyond  L',  A  changes  sign,  being  still  nu- 
merically large.  As  L  continues  to  rotate,  A  increases  alge- 
braically through  negative  values.  Finally,  when  L  has  again 
become  parallel  to  the  axis  of  x,  A  has  increased  algebraically 
through  all  negative  values  and  becomes  again  zero. 


COORDINATES.      CURVES  AND  EQUATIONS        13 


FIG.  6 


When  L  is  in  the  position  of  L',  0  is  90°  and  tan  0  =  X  is 
undefined,  that  is,  has  no  value.  Hence  L'  has  no  slope.  One 
often  sees  the  expression :  tan  90°  =  oo,  and,  in  accordance  with 
it,  one  might  write  here,  A  =  oo .  This  does  not  mean  that  L'  has 
a  slope,  which  is  infinite,  for  "  infinity  "  is  not  a  number.  It 
is  merely  a  brief  and 
symbolic  way  of  describ- 
ing the  behavior  of  X  for 
a  line  L,  near  to,  but  not 
coincident  with  L' ;  it 
says  that  for  such  a  line 
X  is  numerically  very 
large  ;  and  further  that, 
when  the  line  L  ap- 
proaches L'  as  its  limit, 
X  increases  numerically 
without  limit,  —  that  is, 
increases  numerically  be- 
yond any  preassigned  number,  as  10,000,000  or  10,000,000 !,  and 
stays  numerically  above  it. 

The  Angle  0.  In  measuring  the  angle  from  one  line  to  an- 
other, it  is  essential,  first  of  all,  to  agree  on  which  direction 
of  rotation  shall  be  considered  as  positive.  We  shall  take 
always  as  the  positive  direction  of  rotation  that  from  the  posi- 
tive axis  of  x  to  the  positive  axis  of  y ;  so  that  the  angle  from 
the  positive  axis  of  x  to  the  positive  axis  of  y  is  -f  90°,  and 
not  -  90°. 

The  complete  definition  of  0  is,  then,  as  follows :  The  slope- 
angle  0  of  a  line  is  the  angle  from  the  positive  axis  of  x  to  the 
direction  of  the  line.  There  are  in  general  two  positive  values 
for  0  less  than  360° ;  if  the  smaller  of  them  is  denoted  by  0,  the 
other  is  180°  +  0.  Which  of  these  angles  is  chosen  is  imma- 
terial, since  tan  (180°  +  0)  =  tan  0 ;  this  result  is  in  agreement 
with  the  previous  one,  to  the  effect  that  the  slope  pertains  to 
the  undirected  line  without  regard  to  a  sense  on  it. 


14  ANALYTIC  GEOMETRY 

The  student  should  now  draw  a  variety  of  lines,  indicating 
for  each  the  angle  0,  and  assure  himself  that  the  deduction  of 
formula  (3)  holds,  not  merely  when  the  quantities  052  —  x1  and 
#2  —  y\  are  positive,  but  also  when  one  or  both  are  negative. 

Hight-Handed  and  Left-Handed  Coordinate  Systems.     For  the 
choice  of  axes  in  Fig.  1,  the  positive  direction  for  angles  is 
the  counter-clockwise  direction.     But  for 

x    such   a    choice    as    is    indicated    in    the 

present  figure,  —  a  choice  equally  legiti- 
mate, —  it  is  the  clockwise  sense  which  is 
positive. 

-.  The   above    formulas    apply   to   either 

FKJ.  7  system    of    axes.     The    first    system    is 

called  a  right-handed  system ;  the  other, 
a  left-handed  system.  We  shall  ordinarily  use  a  right-handed 
system. 

PROBLEM.  To  draw  a  line  through  a  given  point  having  a 
given  slope.  In  practice,  this  problem  is  usually  to  be  solved 
on  squared  paper.  The  solution  will  be  sufficiently  clearly 
indicated  by  an  example  or  two. 

Example  1.  To  draw  a  line  through  the  point  (—2,  3)  hav- 
ing the  slope  —  4. 

Proceed  along  the  parallel  to  the  a>-axis  through  the  given 
point  by  any  convenient  distance,  as  1  unit,  toward  the  left.* 
Then  go  up  the  'line  through  this  point,  parallel  to  the  ?/-axis, 
by  4  times  the  former  distance,  —  here,  4  units.  Thus,  a  sec- 
ond point  on  the  desired  line  is  determined,  and  the  line  can 
now  be  drawn  with  a  ruler. 

If  the  given  point  lay  near  the  edge  of  the  paper,  so  that 
the  above  construction  is  inconvenient,  it  will  do  just  as  well 
to  proceed  from  the  first  point  toward  the  right  by  1  unit,  and 
then  down  by  four  units. 

*  The  student  will  follow  these  constructions  step  by  step  on  a  piece 
of  squared  paper. 


COORDINATES.   CURVES  AND  EQUATIONS    15 

Example  2.  To  draw  a  line  through  the  point  (1.32,  2.78) 
having  the  slope  .6541. 

Here,  it  is  clear  that  we  cannot  draw  accurately  enough  to 
be  able  to  use  the  last  significant  figure  of  the  given  slope. 
Open  the  compasses  to  span  10  cni.  (if  the  squared  paper  is 
ruled  to  cm.)  and  lay  off  a  distance  of  10  cm.  to  the  right  on  a 
parallel  to  the  ic-axis  through  the  given  point.  This  parallel 
need  not  actually  be  drawn.  Its  intersection,  Q,  with  the  cir- 
cular arc  is  all  that  counts,  and  this  point,  Q,  can  be  estimated 
and  marked.  Its  distance  above  the  axis  of  x  will  be  2  cm. 
and  7.8  mm.  The  error  of  drawing  will  be  of  the  order  of  the 
last  significant  figure,  namely,  more  than  T^  mm.  and  less  than 
.5  mm. 

Next,  open  the  compasses  to  span  6  cm.  and  5.4  mm.  Put 
the  point  of  the  compasses  on  Q,  and  lay  off  the  above  dis- 
tance, 6.54  cm.,  on  a  parallel  through  Q  to  the  y-axis  and 
above  Q.  The  point  R,  thus  found,  will  be  a  second  point  on 
the  desired  line,  which  now  can  be  drawn. 

EXERCISES 

1.  The  points  Pj,  P2,  P3,  with  the  coordinates  (2,  5),  (7,  3), 
( —  3,  7)  respectively,  lie  on  a  line.     Show  that  the  value  for 
the  slope  of  the  line  as  given  by  equation  (3)  is  the  same,  no 
matter  which  two  of  the  three  points  are  used  in  obtaining  it. 

2.  Find  the  slopes  of  the  sides  of  the  triangle  of  Ex.  2,  §  3. 

3.  Find  the  angles  which  the  sides  of  that  triangle  make 
with  the  axes,  and  hence  determine  the  angles  of  the  triangle. 

4.  Show  that  the  points  (-  2,  -  3),  (5,  -  4),  (4,  1),  (—  3,  2) 
are  the  vertices  of  a  parallelogram. 

5.  Draw   a   line   through   the   point   (1,    —2)   having   the 
slope  3. 

6.  Draw  a  line  through  the  point    (—2,   —  1)  having   the 
slope  —  li. 

7.  Draw  a  line  through  the  point  (—1.32,  0.14)  having  the 
slope  -.2688. 


16  ANALYTIC  GEOMETRY      . 

5.   Mid-Point  of  a  Line-Segment.     Let  P1}  with  the  coordi- 
nates (a,1!,  2/i),  and  P2 :  (^  y-i)  ^e  the  extremities  of  a  line-seg- 
ment.    It  is  desired  to  find  the 
Ja  •  (^2^2)    coordinates  of  the  point  P  which 
bisects  P^. 

Let  the  coordinates  of  P  be 
(x,  y).     It  is  evident  that   the 
directed    line-segment    PiP    is 
— x     equal   to  the  directed   line-seg- 


0 

FIG.  8  ment  PP2.     Hence  the  projec- 

tion of  PjP  on  the  axis  of  x,  or 
x  —  ojj,  must  equal  the  projection  of  PP2  on  that  axis,  or  a^  —  x  : 

x  —  x1  =  x2  —  x.      ";.-' 
Hence 


Similar  considerations  apply  to  the  projections  on  the  axis 
of  y,  and  consequently 


2 

We  have  thus  obtained  the  following  result  :  The  coordinates 
(x,  y~)  of  the  point  Ptohich  bisects  the  line-segment  PiP2  are  given 
by  the  equations  : 


EXERCISES 

1.  Determine  the  coordinates  of  the  mid-point  of  each  of 
the  line-segments  given  by  the  pairs  of  points  in  Ex.  1,  §  3. 
Draw  figures  and  check  your  answers. 

2.  Find  the  mid-points  of  the  sides  of  the  triangle  mentioned 
in  Ex.  2,  §  3,  and  check  by  a  figure. 

3.  Determine  the  coordinates  of  the  mid-point  of  the  line 
joining  the  points  (a  +  b,  a)  and  (a  —  6,  6). 

4.  Show  that  the  diagonals  of  the  parallelogram  of  Ex.  4, 
§  4  bisect  each  other. 


COORDINATES.   CURVES  AND  EQUATIONS    17 

6.  Division  of  a  Line-Segment  in  Any  Ratio.*  Let  it  be  re- 
quired to  find  the  coordinates  (x,  y}  of  the  point  P  which 
divides  the  line-segment  PiP2  in  an  arbitrary  ratio,  W1/m2  :  f 

PlP_m1 

PP2  ~  mz  ' 

Obviously  the  projections  of  PjP  and  PP2  on  the  axis  of  x 
must  be  in  the  same  ratio,  m^m^,  and  hence 

x  —  Xj  _  Wi 
#2  —  x     m-i 

On  solving  this  equation  for  x,  it  is  found  that 


Similar  considerations,  applied  to  the  projections  on  the 
axis  of  y,  lead  to  the  corresponding  formula  for  y,  and  thus 
the  coordinates  of  P  are  shown  to  be  the  following  : 


x  _. 


If  ?^i  and  m2  are  equal,  these  formulas  reduce  to  those  of 
§5. 

External  Division.  It  is  also  possible  to  find  a  point  P  on 
the  indefinite  straight  line  through  Pl  and  P2  and  lying  outside 
the  line-segment  P]P2,  which  makes 

PLP=m1 
P2P     w^" 

where  m1  and  m2  are  any  two  unequal  positive  numbers.     Here, 
x1  —  x  _  mi 
x%  —  x     m2 

•This  paragraph  may  well  be  omitted  till  the  results  are  needed  in 
later  work. 

t  The  given  numbers  mi  and  m2  may  be  precisely  the  lengths  PiP 
and  PP2  ;  but  in  general  they  are  merely  proportional  respectively  to 
them,  i.e.  they  are  these  lengths,  each  multiplied  by  the  same  positive  or 
negative  number. 


18  ANALYTIC  GEOMETRY 

On  solving  this  equation  for  x  and  the  corresponding  one  for 
y,  we  find,  as  the  coordinates  of  the  point  P,  the  following  :. 

/2\  x 


m2  —  wii  ra2  —  mx 

The  point  P  is  here  said  to  divide  the  line  PiP2  externally  in 
the  ratio  m1/m2  ;  and,  in  distinction,  the  division  in  the  earlier 
case  is  called  internal  division.  Both  formulas,  (1)  and  (2), 
can  be  written  in  the  form  (1)  if  one  cares  to  consider  external 
division  as  represented  by  a  negative  ratio,  ml/mz,  where,  then, 
one  of  the  numbers  m^  m2  is  positive,  the  other,  negative. 

EXERCISES 

1.  Find  the  coordinates  of  the  point  on   the   line-segment 
joining  (—1,  2)  with  (5,  —  4)  which  is  twice  as  far  from  the 
first  point  as  from  the  second.     Draw  the  figure   accurately 
and  verify. 

2.  Find  the  point  on  the  line  through  the  points  given  in 
the  preceding  problem,  which  is  outside  of  the  line-segment 
bounded  by  them  and  is  twice  as  far  from  the  first  point  as 
from  the  second. 

3.  Find  the  point  which  divides  internally  the  line-segment 
bounded  by  the  points  (3,  8)  and  (—  6,  2)  in  the  ratio  1  :  5,  and 
lies  nearer  the  first  of  these  points. 

4.  The  same  question  for  external  division. 

y^  7.  Curve  Pitting.  Equation  of  a  Curve.  Since  the  subject 
of  graphs  is  now  very  generally  taught  in  the  school  course 
in  Algebra,  most  students  will  already  have  met  some  of  the 
topics  taken  up  on  the  foregoing  pages,  and  moreover  they 
will  have  plotted  numerous  simple  curves  on  squared  paper 
from  given  equations.  Thus,  in  particular,  they  will  be  famil- 
iar with  the  fact  that  all  the  points  whose  coordinates  satisfy 
a  linear  equation,  i.e.  an  equation  of  the  first  degree,  like 

(1)  2x-3y-l  =  0, 


COORDINATES.   CURVES  AND  EQUATIONS    19 


lie  on  a  straight  Hue,  though  they  may  never  have  seen  a 
formal  proof. 

A  number  of  points,  whose  coordinates  satisfy  equation  (1), 
can  be  determined  by  giving  to  x  simple  values,  computing 
the  corresponding  values  of  y  from  (1),  and  then  plotting  the 
points  (x,  y).  Thus 

if  x  =  0,      y  =  —  ^,  and  the  point  is  (0,  —  £)  , 
if  x  =  1,      y  =  i,      and  the  point  is  (1,  -j) ; 
if  x  =  2,      y  =  1,      and  the  point  is  (2,  1) ; 
if  x  =  —  1,  y  =  —  1,  and  the  point  is  (—  1,  —  1); 
etc. 

Of  course,  if  it  is  known  that  (1)  represents  a  straight  line, 
—  i.e.  that  all  the  points  whose 
coordinates  satisfy  (1)  lie  on  a 
straight  line,  —  it  is  sufficient 
to  determine  two  points  as  above, 
and  then  to  draw  the  line  O 

through  them. 

This  process  of  determining  a 
large  number  of  points  whose 
coordinates  satisfy  a  given  equa-  FIG.  9 

tion  and  then  passing  a  smooth 

curve  through  them  is  known  as  "  plotting  a  curve  *  from  its 
equation." 

The  mathematical  curve  f  defined  by  an  equation  in  x  and 
y  consists  of  all  those  points  and  only  those  points  whose  coordi- 
nates, when  substituted  for  x  and  y  in  the  equation,  satisfy  it. 

Suppose,  for  example,  that  the  equation  is 
(2)  y  =  x\ 

The  point  (2,  4)  lies  on  the  curve  defined  by  (2),  because,  when 

*  In  Analytic  Geometry  the  term  curve  includes  straight  lines  as  well 
as  crooked  curves. 

t  This  curve  is  sometimes  called  the  locus  of  the  equation. 


20  ANALYTIC  GEOMETRY 

x  is  set  equal  to  2  and  y  is  set  equal  to  4  in  (2),  the  resulting 
equation, 

4  =  4, 

is  true.     We  say,  equation  (2)  is  satisfied  by  the  coordinates  of 
the  point  (2,  4),  or  that  the  point  (2,  4)  lies  on  the  curve  (2) 
On  the  other  hand,  the  point  (—  1,  2),  for  example,  does  not 
lie  on  the  curve  denned  by  (2).     For,  if  we  set  x  =  —  1  and 
y  =  2,  equation  (2)  becomes 

2  =  1. 

This  is  not  a  true  equation;  i.e.  equation  (2)  is  not  satisfied 
by  the  coordinates  of  the  point  (—  1,  2),  and  so  this  point  does 
not  lie  on  the  curve  (2). 

Equation  of  a  Curve.  A  curve  may  be  determined  by  simple 
geometric  conditions ;  as,  for  example,  that  all  of  its  points 
be  at  a  distance  of  2  units  from  the  origin.  This  is  a  circle 
with  its  center  at  the  origin  and  having  a  radius  of  length  2. 

It  is  easy  to  state  analytically  the  condition  which  the  coor- 
dinates of  any  point  (x,  y)  on  the  circle  must  satisfy.  Since 
by  §  3  the  distance  of  any  point  (x,  y)  from  the  origin  is 


the  condition  that  (x,  y)  be  a  point  of  the  curve  is  clearly  this, 
that 


or  that 

(3)  0^  +  2/2  =  4. 

Equation  (3)  is  called  the  equation  of  the  curve  in  question. 

The  equation  of  a  curve  is  an  equation  in  x  and  y  which  is 
satisfied  by  the  coordinates  of  every  point  of  the  curve,  and  by 
the  coordinates  of  no  other  point. 

In  this  book  we  shall  be  engaged  for  the  most  part  in  find- 
ing the  equations  which  represent  the  simpler  and  more  im- 
portant curves,  and  in  discovering  and  proving,  from  these 
equations,  properties  of  the  curves. 


COORDINATES.      CURVES  AND  EQUATIONS        21 

Nevertheless,  the  student  should  at  the  outset  have  clearly 
in  mind  the  fact  that  any  equation  between  x  and  y,  like 

y  =  x3,  y  =  \ogx,  y=  sin  x, 

represents  a  perfectly  definite  mathematical  curve,  which  he 
can  plot  on  paper.  Moreover,  he  is  in  a  position  to  determine 
whether,  in  the  case  of  a  chosen  one  of  these  curves,  a  given 
point  lies  on  it.  He  will  find  it  desirable  to  plot  afresh  a  few 
simple  curves,  and  to  test  his  understanding  of  other  matters 
taken  up  in  this  paragraph  by  answering  the  questions  in  the 
following  exercises. 

EXERCISES 

1.  What  does  each  of  the  following  equations  represent? 
Draw  a  graph  in  each  case. 

(o)  a?  =  2;  (c)  a?  — y  =  0;  (e)  2x  -  3y  +  6  =  0; 

(6)  2y  +  3  =  0;      (d)  2# -f- 5y  =  0 ;      (y)  5x  -j-  Sy  —  4  =  0. 

Plot  the  following  curves  on  squared  paper. 

2.  y  —  x\ 

Take  2  cm.  or  1  in.  as  the  unit  of  length.     Use  a  table  of 
squares. 

3.  y1  =  x. 

Take  the  same  unit  as  in  question  2  and  use  a  table  of  square 
roots. 

4.  Show  that,  when  one  of  the  curves  of  Exs.  2  and  3  has 
been  plotted  from  the  tables,  the  other  can  be  plotted  from 
the  first  without  the  tables. 

Work  the  corresponding  exercises  for  the  following  curves. 

5.  y  =  x3.  6.  y  =  Va/.  7.   y1  =  x3. 
9.   Plot  the  curve 


from  a  table  of  logarithms  for  values  of  x  from  1  to  10,  taking 
1  cm.  as  the  unit. 


22  ANALYTIC   GEOMETRY 

10.   Which  of  the  straight  lines  of  Ex.  1  go  through  the 
origin  ? 
I/  11.    Show  that  the  curve 

(a)  y  •=•  sin  x 

goes  through  the  origin. 
Do  the  curves 

(6)  y  =  tan  x,  (c)  y  =  cos  x, 

go  through  the  origin  ? 

12.   Do  the  following  points  lie  on  the  curve 

xy  =  l? 
(a)  (-!,-!);  (6)  (-1,1);  (c)  (f,  f  )  ; 

(rf)(-i-f);         00  (i,  -2);         (/)(o,i). 

(  13)   Find  the  equations  of  the  following  curves. 

(a)  The  line  parallel  to  the  axis  of  x  and  8  units  above  it. 

(6)  The  line  parallel  to  the  axis  of  y  and  If  units  to  the 
left  of  it. 

(c)  The  line  bisecting  the  angle  between  the  positive  axis 
of  y  and  the  negative  axis  of  x. 

(d)  The  circle,  center  in  the  origin,  radius  p. 

(e)  The  circle,  center  in  the  point  (1,  2),  radius  3. 

Ans.     z-l 


8.  Points  of  Intersection  of  Two  Curves.  Consider,  for  ex- 
ample, the  problem  of  finding  the  point  of  intersection  of  the 
lines 

L:  2x-3y  =  ±, 

L':  3z  +  4t/  =  -ll. 

Let  (a?!,  yi)  be  the  coordinates  of  this  unknown  point,  Pj. 
Any  point  P,  with  the  coordinates  (x,  y),  which  lies  on  L,  has 
its  x  and  y  satisfying  the  first  of  the  above  equations.  Hence, 
in  particular,  since  PT  lies  on  L,  xl  and  yt  must  satisfy  that 
equation,  or 
(1)  2^-3^  =  4. 


COORDINATES.      CURVES  AND  EQUATIONS        23 

Similarly,  a  point  P :  (x,  y),  which  lies  on  L',  has  its  x  and 
y  satisfying  the   second   of   the   above   equations.     Hence,  in 
particular,  since  PI  lies  on  L',  x^  and  yl  must  satisfy  that  equa- 
tion, or 
(2)  3^+4^  =  -11. 

Thus  it  appears  that  the  two  unknown  quantities,  x±  and  y1} 
satisfy  the  two  simultaneous  equations,  (1)  and  (2).  Hence 
these  equations  are  to  be  solved  as  simultaneous  by  the 
methods  of  Algebra. 

2x1  —  3y1=       4,         4 

To  do  this,  eliminate  yl  by  multiplying  the  first  equation 
through  by  4,  the  second  by  3,  and  then  adding : 

17  #!  =  —  17,         or         xl  =  —  1. 

On  substituting  this  value  of  xl  in  either  equation  (1)  or  (2), 
the  value  of  yi  is  found  to  be :  y\  =  —  2.  Hence  Pl  has  the 
coordinates  (—  1,  —2). 

The  equations  (1)  and  (2)  are  the  same,  except  for  the  sub- 
scripts, as  the  equations  of  the  given  lines,  L  and  L'.  Hence 
we  may  say  :  To  find  the  coordinates  of  the  point  of  intersection 
of  two  lines  given  by  their  equations,  solve  the  latter  as  simul- 
taneous equations  in  the  unknoivn  quantities,  x  and  y,  by  the 
methods  of  Elementary  Algebra. 

The  generalization  to  the  case  of  any  two  curves  given  by 
their  equations  is  obvious.  The  equations  are  to  be  regarded 
as  simultaneous  equations  between  the  unknown  quantities,  x  and 
y,  and  solved  as  such. 

The  student  should  observe  that  the  letters  "  x "  and  "  y " 
have  totally  different  meanings  when  they  appear  as  the  co- 
ordinates of  a  variable  point  in  the  equation  of  a  curve,  and  when 
they  represent  unknown  quantities  in  a  pair  of  simultaneous 
equations.  In  the  first  case,  they  are  variables,  and  a  pair  of 
values,  (x,  y),  which  satisfy  equation  L  will  not,  in  general, 
satisfy  L'  In  the  second  case,  x  and  y  are  constants,  the 


24  ANALYTIC   GEOMETRY 

coordinates  of  a  single  point,  or   of   several   points  ;    but   of 
isolated  and  not  variable  points. 

EXERCISES 

Determine  the  points  of  intersection  of  the  following  curves. 
Check  your  results  by  plotting  the  curves  and  reading  off  as 
accurately  as  possible  the  coordinates  of  the  points  of 
intersection. 

1.  The  straight  lines  (a)  and  (<f)  of  Ex.  1,  §  7. 

2.  The  straight  lines  (c)  and  (e)  of  Ex.  1,  §  7. 

3.  The  straight  lines  (e)  and  (/)  of  Ex.  1,  §  7. 

f  yi=  4#, 

+  y=3.  Ans.   (1,  2),  (9,  -6). 


xy  =  6.  }  a;  +  y  =  0. 

f 
7. 


=  144. 

z2.i^/2  =  2, 

^»«-  (1,1),  (-1,  -1). 

10.    Show  that  the  curves 


intersect  in  the  point  (1,  0). 
11.    Show  that  the  curves 


intersect  in  the  point  (4,  3),  and  also  in  (—  4,  —  3). 

^T^XERCJSES  ON  CHAPTER  I 

X-Show  that  the  points  (2,  0),  (0,  2),  (1  +  V3,  1  +  V3)  are 
the  vertices  of  an  equilateral  triangle. 

.Xfc-  Prove  that  the  triangle  with  vertices  in  the  points  (1,  8), 
(3,  2),  (9,  4)  is  an  isosceles  right  triangle. 


COORDINATES.      CURVES  AND  EQUATIONS        25 

Show   that   the   points    (- 1,   2),    (4,   10),   (2,  3),  and 
— 5)  are  the  vertices  of  a  parallelogram. 

Given  the  points  A,  B,  C  with  coordinates  (—7,  —2), 
,  0),  (5,  3).     By  proving  that 

AB  +  BC=AC, 

show  that  the  three  points  lie  on  a  line. 

vL    Show  that  the  three  points  of  the  previous  problem  lie 
on  a  line  by  proving  that  AB  and  AC  have  the  same  slope. 

— vA.   Prove  that  the  two  points  (5,  3)  and  (—  10,  —  6)  lie  on 
a  Itiie  with  the  origin. 

a-^t*   Prove  that  the  two  points  (o^,  3^),  (cc2,  3/2)  lie  on  a  line 
witn  the  origin  when,  and  only  when,  their   coordinates  are 

proportional : 

X1:y1  =  x2:  y2. 

"^    Determine  the  point  on  the  axis  of  x  which  is  equidis- 
tant from  the  two  points  (3,  4),  (—2,  6). 

9.  If  (3,  2)  and  (—3,  2)  are  two  vertices  of  an  equilateral 
triangle  which  contains  within  it  the  origin,  what  are  the  co- 
ordinates of  the  third  vertex  ? 

10.  If  (3,  —  1),  (—4,   —3),  (1,  5)  are  three  vertices  of  a 
parallelogram  and  the  fourth  lies  in  the  first  quadrant,  find 
the  coordinates  of  the  fourth.  Ans.    (8,  7). 

11.  If  P  is  the  mid-point  of  the  segment  PiP2,  and  P  and 
Pl  have  coordinates  (8,  17),  (—5,  —  3)  respectively,  what  are 
the  coordinates  of  P2  ? 

12.  If  P  divides  the  segment  PiP2  in  the  ratio  2  : 1,  and  Pt 
and  P  have  coordinates  (3,  8)  and  (1,  12)  respectively,  deter- 
mine the  coordinates  of  P2.  Ans.    (0,  14). 

13.  Find  the  ratio  in  which  the  point  B  of  Ex.  4  divides 
the  segment  AC  of  that  exercise.  Ans.   2  :  3. 

14.  A  point  with  the  abscissa  6  lies  on  the  line  joining  the 
two  points  (2,  5),  (8,  2).     Find  its  ordinate. 


26  ANALYTIC    GEOMETRY 

Suggestion.     Determine  the  ratio  in  which  the  point  divides 
the  line-segment  between  the  two  given  points. 

"l^.  Prove  that  the  sum  of  the  squares  of  the  distances  of 
any  point  in  the  plane  of  a  given  rectangle  to  two  opposite 
vertices  equals  the  sum  of  the  squares  of  the  distances  from  it 
to  the  two  other  vertices. 

Suggestion.     Choose  the  axes  of  coordinates  skillfully. 

16.   If  D  is  the  mid-point  of  the  side  BO  of  a  triangle  ABC, 
prove  that 

ABZ  +  AC2  =  2  AD2  +  2  BD2. 


Show  that  the  lines  joining  the  mid-points  of  opposite 
sides  of  a  quadrilateral  bisect  each  other. 

*"^S,  Prove  that  the  lines  joining  the  mid-points  of  adjacent 
sides  of  a  quadrilateral  form  a  parallelogram. 

19.  Prove  that,  if  the  diagonals  of  a  parallelogram  are  equal, 
the  parallelogram  is  a  rectangle. 

20.  If  two  medians  of  a  triangle  are  equal,  show  that  the 
triangle  is  isosceles. 


CHAPTER   II 


THE   STRAIGHT   LINE 


1.  Equation  of  Line  through  Two  Points.  Let  P]  :  (xl}  t/x) 
and  P2 :  (x2,  y2)  be  two  given  points,  and  let  it  be  required  to 
find  the  equation  of  the  line  through 
them. 

The  slope  of  the  line,  by  Ch.  I, 
§  4,  is 


FIG. 


Let  P,  with   the   coordinates   (x,  y\ 

be  any  point  on  the  line  other  than  Pv 

the  line  is  also  given  by 

y  —  yiT 


Then  the  slope  of 


X  — 


Hence 
(1) 


X  — 


—  Xl 


Conversely,  if  P  :  (x,  y)  is  any  point  whose  coordinates 
satisfy  equation  (1),  this  equation  then  says  that  the  slope  of 
the  line  PjP  is  the  same  as  the  slope  of  the  line  PiP2  and 
hence  that  P  lies  on  the  line  PiP2. 

A  more  desirable  form  of  equation  (1)  is  obtained  by  multi- 
plying each  side  by  (x  —  xl)/(y2  —  y^.  We  then  have  : 

(I)  x  -xi  =  V  —  yi  . 


Equation  (I)  is  satisfied  by  the  coordinates  of  those  points  and 
only  those  points  which  lie  on  the  line  PiP2.     Consequently, 

27 


28  ANALYTIC   GEOMETRY 

by  Ch.  1,  §  7,  (I)  is  the  equation  of  the  line  through  the  two  given 
points. 

Example  1.     Find  the  equation  of  the   line  which   passes 
through  the  points  (1,  —  2)  and  (—  3,  4). 

Here 

<KI  =  1,     2/i  =  —  2         and         x^  =  —  3,     y2  =  4. 
By  (I)  the  equation  of  the  line  is 


-3-1     4  -(-2)'  -4  "      6 

On  clearing  of  fractions  and  reducing,  the  equation  becomes 
3x  +  2y  +  1  =  0. 

Let  the  student  show  that,  if  (xt,  y^)  had  been  taken  as 
(—  3,  4)  and  (a^,  y%)  as  (1,  —  2),  the  same  equation  would  have 
resulted. 

Example  2.  Find  the  equation  of  the  line  passing  through 
the  origin  and  the  point  (a,  6). 

Here,  fa,  yl)  =  (0,  0)  and  (a^,  ?/2)  =  (a,  6),  and  (I)  becomes 

-  =  -  ,         or         foe  —  ay  =  0. 
a      6 

Lines  Parallel  to  the  Axes.  In  deducing  (I)  we  tacitly  as- 
sumed that 

2/2  —  2/1^0         and         x*  —  x^  ^  0  ; 

for  otherwise  we   could    not    have   divided    by   these   quan- 
tities. 

If  2/2  —  2/i  =  0,  the  line  is  parallel  to  the  axis  of  x.  Its 
equation  is,  then,  obviously 

(2)  y  =  y* 

Similarly,  if  x%  —  x{  =  0,  the  line  is  parallel  to  the  axis  of  y 
and  has  the  equation 

(3)  x  =  xi. 

These  two  special  cases  are  not  "included  in  the  result  eui- 


THE   STRAIGHT   LINE  29 

bodied  in  equation  (I).     We  see,  however,  that   they  are  so 
simple,  that  they  can  be  dealt  with  directly.* 

Example  3.     Find  the  equation  of  the  line  passing  through 
the  two  points  (—5,  1)  and  (—5,  8). 

It  is  clear  from  the  figure  that  this  line  is  parallel  to  the 
axis  of  y  and  5  units  distant  from  it  to  the 
left.  Accordingly,  the  abscissa  of  every  point 
on  it  is  —  5 ;  conversely,  every  point  whose 
abscissa  is  —  5  lies  on  it.  Therefore,  its  equa- 
tion  is 


o1 
x  =  —  5,         or         x  +  5  =  0.  FIG.  2 

EXERCISES  t 
Draw  the  following  lines  and  find  their  equations. 

1.  Through  (1,  1)  and  (3,  4).  Ans.    3x-2y  —  l  =  0. 

2.  Through  (5,  3)  and  (-  8,  6). 

3'.  Through  (0,  -  5)  and  (—  2,  0).     Ans.   5x+2y  +  W  =  0. 

4.  Through  the  origin  and  (—1,  2). 

5.  Through  the  origin  and  (—  2,  —  3). 

6.  Through  (2,  -  3)  and  (-  4,  -  3).  Ans.   y  +  3  =  0. 

*  It  is  not  difficult  to  replace  (I)  by  an  equation  which  holds  in  all 
cases,  —  namely,  the  following  : 

(I')  (2/2  -  2/i)  O  -  Xi)  =  (x2  -  xO  (y  -  yi). 

We  prefer,  however,  the  original  form  (I).  For  (I)  is  more  compact 
and  easier  to  remember,  and  the  special  cases  not  included  in  it  are  best 
handled  without  a  formula. 

t  In  substituting  numerical  values  for  (xi,  y\)  and  (x2,  3/2)  in  (I),  the 
student  will  do  well  to  begin  with  a  framework  of  the  form 

x  -         y  — 

i 

and  then  fill  in  each  place  in  which  Xi  occurs  ;  next,  each  place  in  which 
j/i  occurs  ;  and  so  on.  When  x\  or  y\  is  negative,  substitute  it  first  in 
parentheses  ;  thus,  if  xt  =  —  3,  begin  by  writing 

-" 


30  ANALYTIC   GEOMETRY 

7.  Through  (0,  8)  and  (0,  -  56). 

8.  Through  (5,  3)  and  parallel  to  the  axis  of  y. 

9.  Through  (5,  3)  and  parallel  to  the  axis  of  x. 

10.  Through  (a,  6)  and  (b,  a).  Ans.   x  +  y  =  a  +  b 

11.  Through  (a,  0)  and  (0,  6).  A^    x  +  y=l 

a     b 

2.  One  Point  and  the  Slope  Given.  Let  it  be  required  to  find 
the  equation  of  the  line  which  passes  through  a  given  point 
PI  '•  (xi>  y\)  and  has  a  given  slope,  A. 

If  P  :  (x,  y)  be  any  second  point  on  the  line,  the  slope  of  the 
line  will  be,  by  Ch.  I,  §  4, 

y  -  yi  . 

X—  Xi 

« 

But  the  slope  of  the  line  is  given  as  A.     Hence 


X—  Xl 

or 

(II)  y  -  yi  =  A(a?  —  »i). 

The  student  can  now  show,  conversely,  that  any  point,  whose 
coordinates  (x,  y~)  satisfy  (II),  lies  on  the  given  line.  Hence 
(II)  is  the  equation  of  the  line  passing  through  the  given  point 
and  having  the  given  slope. 

Example.  Find  the  equation  of  the  line  which  goes  through 
the  point  (2,  —  3)  and  makes  an  angle  of  135°  with  the  posi- 
tive axis  of  x. 

Here,  X  =  —  1  and  (xl}  yl)  =  (2,  —  3),  and  hence,  by  (II),  the 
equation  of  the  line  is 

</  +  3  =  -l(z-2), 
or  x  +  y  +  1  =  0. 

Slope-  Intercept  Form  of  Equation.  It  is  frequently  conven- 
ient to  determine  a  line  by  its  slope  X,  and  .the  ^-coordinate 
of  the  point  in  which  it  cuts  the  axis  of  y. 


THE   STRAIGHT   LINE 


31 


Here,  Xi  =  0 ;  and,  if  we  denote  yt  by  the  letter  b,  (II)  be- 
comes 
(III)  y  =  \x  +  b. 

This  is  known  as  the  slope-intercept  form  of   the   equation 
of    a    straight    line ;    b    is    known    as 
the   intercept  of   the   line   on   the  axis 
of  y. 


Example.     Find  the  equation  of  the    ^ 

line  which  makes  an  angle  of  60°  with    ! 

the  axis  of  x  and  whose  intercept  on  the 
axis  of  y  is  —  2. 

Since  A  =  V3  and  b  =  —  2,  the  equation  is 


(0,6) 


FIG.  3 


EXERCISES 

Draw  the  following  lines  and  find  their  equations. 
1.    Through  (—  4,  5)  and  with  slope  —  2. 

Ans.   2x  +  y+3  =  0. 
^2.   Through  (3,  0)  and  with  slope  f. 

3.  Through  (f ,  —  i)  and  with  slope  —  f . 

4.  Through  the  origin  and  making  an  angle  of  60°  with  the 
axis  of  x. 

5.  Through  (—4,  0)  and  making  an  angle  of  45°  with  the 
axis  of  y. 

6.  With  intercept  1  on  the  axis  of  y  and  with  slope  —  f . 

Ans.   3x  -\-2y  —  2  =  0. 

//\1.   With  intercept  ^  on  the  axis  of  y  and  making  an  angle 
of  30°  with  the  axis  of  x. 

8.  With  slope  —  1  and  intercept  —  c  on  the  axis  of  y. 

9.  With  slope  a/b  and  intercept  6  on  the  axis  of  y. 


I.  The  General  Equation  of  the  First  Degree.     Let  there  be 
given  an  arbitrary  line  of  the  plane.     If  the  line  is  parallel 


32  ANALYTIC   GEOMETRY 

to  neither  axis,  its  equation  is  of  the  form  (I),  §  1,  —  an  equa- 
tion of  the  first  degree  in  x  and  y.  If  the  line  is  parallel  to 
the  axis  of  re,  its  equation  is  of  the  form  y  =  yl}  —  a  special 
equation  of  the  first  degree  in  x  and  y,  in  which  it  happens 
that  the  term  in  x  is  lacking.  Similarly,  if  the  line  is  parallel 
to  the  axis  of  y,  its  equation  is  of  the  form  x  =  xl}  —  an  equa- 
tion of  the  first  degree  which  lacks  the  term  in  y.  Conse- 
quently, we  can  say  :  The  equation  of  every  straight  line  is  of  the 
first  degree  in  x  and  y. 

Given,  conversely,  the  general  equation  of  the  first  degree  in 
x  and  y,  namely 

(1)  Ax  +  By+C=0, 

where  A,  B,  C  are  any  three  constants,  of  which  A  and  B  are 
not  both  zero  ;  *  this  equation  represents  always  a  straight  line. 
The   Case  B^Q.    In  general,  B  will  not  be  zero  and  we 
can  divide  equation  (1)  through  by  it: 


and  then  solve  for  y  : 

v  —  —  —  x  —  —  • 
B        B 

But  this  equation  is  precisely  of  the  form  (III),  §  2,  where 

A  ,        .~C 

\  —  --  ,  o  =  -  --- 

B  B 

Therefore,  it  represents  a  straight  line  whose  slope  is  —  A/B 
and  whose  intercept  on  the  axis  of  y  is  —  C/B. 

The  Case  B  =  0.     If,  however,  B  is  zero,  the  equation  (1) 
becomes 


Now,  A  cannot  be  zero,  since  the  case  that  both  A  and  B  are 
zero  was  excluded  at  the  outset.  We  can,  therefore,  divide  by 
A  and  then  solve  for  x  : 


*  In  dealing  with  equation  (1),  now  and  henceforth,  we  shall  always 
assume  that  A  and  B  are  not  both  zero.  / 


THE   STRAIGHT   LINE 


33 


This  is  the  equation  of  a  straight  line  parallel  to  the  axis 
of  y,  if  0  =£  0.     If  C  =  0,  it  is  the  equation  of  this  axis. 

This  completes  the  proof  that  every  equation  of  the  first 
degree  represents  a  straight  line.  In  accordance  with  this 
property,  such  an  equation  is  frequently  called 
a  linear  equation. 

Example.     What  line  is  represented  by  the 
equation 


If  we  solve  for  y,  we  obtain 


l\ 

y 

\, 

\ 

\     IT 

\'"i( 

1    1 

FIG.  4 


Hence  the  equation  represents  the  line  of  slope 

—  2  with  intercept  —  -^  on  the  axis  of  y.     From  these  data  we 

may  draw  the  line. 

EXERCISES 

Find  the  slopes  and  the  intercepts  on  the  axis  of  y  of  the 
lines  represented  by  the  following  equations.     Draw  the  lines. 

1.  4oj  +  2y  — 1  =  0.  4.   2x  —  y  = 

2.  7#  +  8?/ +  5  =  0.  5.   y  =  0. 

3.  2x  —  5?/  =  0.  6.   x  =  3  —  y. 

Find  the  slopes  of  each  of  the  following  lines. 

7.  —  x-\-2y  =  7.     Ans.   ^.        11.   2y  —  3  = 

8.  x=y  +  l.  12.   2x  =  3y. 

9.  3  —  2x  =  5y.  13.   x  =  5y  + 
10.    2x-3i/  =  4. 


jt*t 


4.  Intercepts.  In  the  preceding  paragraph  we  learned  to 
plot  the  line  represented  by  a  given  equation,  from  the  values 
of  its  slope  and  its  intercept  on  the  axis  of  y,  as  found  from 
the  equation.  It  is  often  simpler,  however,  in  the  case  of  a 
line  which  cuts  the  axes  in  two  distinct  points,  to  determine 
from  the  equation  the  coordinates  of  these  two  points  and  then 
to  plot  the  points  and  draw  the  line  through  them. 


34  ANALYTIC   GEOMETRY 

The  point  of  intersection  of  a  line,  for  example, 

with  the  axis  of  x  has  its  ^/-coordinate  equal  to  0.  Conse- 
quently, to  find  the  ^-coordinate  of  the  point,  we  have  but  to 
set  y  =  0  in  the  equation  of  the  line  and  solve  for  x.  In  this 
case  we  have,  then, 

2  x  -f-  4  =  0,         or        x  =  —  2. 

Similarly,  the  cc-coordinate  of  the  point  of  intersection  of 
the  line  with  the  axis  of  y  is  0,  and  its  y-coordinate  is  obtained 
by  setting  x  =  0  in  the  equation  of -the  line  and  solving  for  y. 
In  the  present  case  this  gives 

_  3y  +  4  =  0,         or        y  =  -f . 

The  points  of  intersection  of  the  line  (1)  with  the  axes  of 
coordinates  are,  then,  (—2,  0)  and  (0,  -|).     We  now  plot  these 
y  points  and  draw  the  line  through  them. 

<fo,4)  We    recognize    the    number    f    as    the 

intercept  of  the  line  (1)  on  the  axis  of  y ; 
- — t-x  the  number  —  2  we  call  the  intercept  on 
the  axis  of  x.  We  have  plotted  the  line 
(1),  then,  by  finding  its  intercepts. 
In  general,  the  intercept  of  a  line  on  the  axis  of  or  is  the 
^coordinate  of  the  point  in  which  the  line  meets  that  axis.  The 
intercept  on  the  axis  of  y  is  similarly  defined.  These  defini- 
tions admit  of  extension  to  any  curve.  Thus,  the  circle  of  Ch.  I, 
§  7,  has  two  intercepts  on  the  axis  of  x,  namely,  +  2  and  —  2. 
An  axis  or  a  line  parallel  to  an  axis  has  no  intercept  on  that 
axis.  Every  other  line  has  definite  intercepts  on  both  axes, 
and  these  intercepts  determine  the  position  of  the  line  unless 
they  are  both  zero,  that  is,  unless  the  line  goes  through  the 
origin. 

EXERCISES 

Determine  the  intercepts  of  the  following  lines  on  each  of 
the  coordinate  axes,  so  far  as  such  intercepts  exist,  and  draw 
the  lines. 


FIG.  5 


THE    STRAIGHT    LINE  35 

1.  2x  +  3y-6  =  0.  6.   2x  +  3  =  0. 

Ans.   3,  2.  Ans.    —  1^,  none. 

2.  x  —  y  +  1  =  0.  7.   8  —  5y  =  0. 
3x-5y  +  10  =  0.               ^8.   a?=0. 

4.  5a  +  ly  +  13  =  0.  9.   x  +  y  =  a. 

5.  2z-3y  =  0.  >^10.    2ax-3by  =  ab. 

5.  The  Intercept  Form  of  the  Equation  of  a  Line.  Given  a 
line  whose  position  is  determined  by  its  intercepts.  Let  the 
intercept  on  the  axis  of  x  be  cr,  and  let  that  on  the  axis  of  y  be 
b.  To  find  the  equation  of  the  line  in  terms  of  a  and  b. 

Since  one  point  on  this  line  is  (a,  0)  and  a  second  is  (0,  6), 
we  have,  by  (I),  §  1, 

x  —  a  _  y  —  0 
0-a~6-0' 
or 

(IV)  ?  +  £=!. 

a     b 

Only  lines  which  intersect  the  axes  in  two  points  that  are 
distinct  can  have  their  equations  written  in  this  form.  A  line 
through  the  origin  is  an  exception,  because  one  or  both  its 
intercepts  are  zero  and  division  by  zero  is  impossible.  Also  a 
line  parallel  to  an  axis  is  an  exception,  since  it  has  no  inter- 
cept on/ that  axis. 

EXERCISES 

\Find  the  equations  of  the  following  lines. 
*4l.   With  intercepts  5  and  3. 
2.   With  intercepts  —  2-£  and  8. 
With  intercepts  -|  and  —  |. 

4.  The  diagonals  of  a  square  lie  along  the  coordinate  axes, 
and  their  length  is  2  units.  Find  the  equations  of  the  four 
sides  (produced). 

Ans.  x  +  y  =  1 ;     x—  ?/  =  1 ;     —  a;  +  y  =  1 ;     ~  x  —  y  =  1. 


36  ANALYTIC   GEOMETRY 

5.    A  triangle  has  its  vertices  at  the  points  (0,  1),  (—  2,  0), 
(1,  0).     Draw  the  triangle  and  find  the  equations  of  its  sides 
(produced).     Use  formula  (IV),  when  possible. 
^6.    A   triangle  has  its  vertices  at  the  points   (a,  0),  (6,  0), 
(0,  c).     Find  the  equations  of  the  sides  (produced). 

7.  A  line  goes  through  the  origin  and  the  mid-point  of  that 
side  of  the  triangle  of  Ex.  5  which  lies  in  the  first  quadrant. 
Find  its  equation. 

8.  Find  the  equations  of  the  lines  through  the  origin  and 
the.  respective  mid-points  of  the  sides  of  the  triangle  of  Ex.  6. 


Parallel  and  Perpendicular  Lines.  Parallels.  Given  two 
lines  oblique  to  the  axis  of  y,  so  that  both  have  slopes.  The 
lines  are  parallel  if,  and  only  if,  they  have  equal  slopes.  For, 
if  they  are  parallel,  their  slope  angles,  and  hence  their  slopes, 
are  equal  ;  and  conversely. 

Example  1.     To  find  the  equation  of  the  line  through  the 
point  (1,  2)  parallel  to  the  line 
(1)  3x-2y  +  6  =  Q. 

The  slope  of  the  line  (1)  is  f  .  The  required  line  has  the 
same  slope  and  passes  through  the  point  (1,  2).  By  (II),  §  2, 
its  equation  is 

y-2«f(*-l), 
or 


If  the  given  line  is  parallel  to  the  axis  of  y,  it  has  no  slope  and 
hence  the  method  of  Example  1  is  inapplicable.  But  then  the 
required  line  must  also  be  parallel  to  the  axis  of  y  and  its 
equation  can  be  written  down  directly.  For  example,  if  the 
given  line  is  3x  +  8  =  0,  and  there  is  required  the  line  parallel 
to  it  passing  through  the  point  (—8,  2),  it  is  clear  that  the 
required  line  is  parallel  to  the  axis  of  y  and  8  units  to  the  left 
of  it,  and  consequently  has  the  equation  x  =  —  8,  or  x  -f-  8  =  0. 

Perpendiculars.  Given  two  lines  oblique  to  the  axes,  so  that 
both  have  slopes,  neither  of  which  is  zero.  The  lines  are  per- 


THE   STRAIGHT   LINE  37 

pendicular  if,  and  only  if,  their  slopes,  \{  and  A2,  are  negative 
reciprocals  of  one  another  : 

(2)  A2  =  -f,         or         Xi  =  -f,         A^O,  A2  =£  0. 

A!  A2 

For,  if  the  lines  are  perpendicular,  one  of  their  slope  angles, 
61  and  #2>  may  be  taken  as  90°  greater  than  the  other,  viz.  : 

02  =  0j  +  90°, 
and  hence 

Ao  =  tan  02  =  tan  (0t  +  90°)  =  -  cot  ^  = 


tan  Oi         \i 
or 

A.—  i: 

Xi 

Conversely,  if  this  last  equation  is  valid,  the  steps  can  be 
retraced  and  the  lines  shown  to  be  perpendicular  to  each 
other. 

Example  2.  To  find  the  equation  of  the  line  through  the 
point  (1,  2)  perpendicular  to  the  line  (1). 

The  slope  of  (1)  is  f  .  Hence  the  required  line  has  the  slope 
—  |.  We  have,  then,  to  find  the  equation  of  the  line  through 
the  point  (1,  2)  with  slope  —  -|.  By  (II),  §  2,  this  equation  is 


If  the  given  line  is  parallel  to  an  axis,  it  has  no  slope  or  its 
slope  is  zero.  In  either  case,  equation  (2)  and  the  method  of 
Example  2  are  inapplicable.  But  then  th^  required  line  must 
be  parallel  to  the  other  axis  and  it  is  easy  to  write  its  equation. 
Suppose,  for  example,  that  the  given  line  is  2y  —  3=0,  —  a 
line  parallel  to  the  axis  of  x,  —  and  that  the  required  line  per- 
pendicular to  it  is  to  go  through  the  point  (3,  5).  Then  this 
line  must  be  parallel  to  the  axis  of  y  and  at  a  distance  of 
3  units  to  the  right  of  it.  Consequently,  its  equation  is 
x  -  3  =  0. 

The  methods  of  this  paragraph  are  applicable  to  all  problems 


38  ANALYTIC   GEOMETRY 

in  which  it  is  required  to  find  the  equation  of  a  line  which 
passes  through  a  given  point  and  is  parallel,  or  perpendicular, 
to  a  given  line. 

EXERCISES 

In  each  of  the  following  exercises  find  the  equations  of  the 
lines  through  the  given  point  parallel  and  perpendicular  to 
the  given  line. 

Line  Point 

-8y  =  5,  (-1,  -3). 

Ans.    x  —  2y  —  5  =  0;     2x  +  y  +  5  =  Q. 

2.  x-y  =  l,  (0,0). 

3.  5x  +  13y  -  3  =  0,       (2,  -  1). 

4.  3x  +  5y  =0,  (5,0). 

>.   2  x  =  3,  (5,  -6). 

6.    V2y  +  7r  =  0,  (-2,0).     Ans.   y  =  0;     x  +  2  =  0. 

.   1  -  x  =  0,  (0,  *-). 

8.  Find  the  equations  of  the  altitudes  of   the   triangle   of 
§  5,  Ex.  5. 

9.  Find  the  equations  of  the  perpendicular  bisectors  of  the 
sides  of  the  triangle  of  §  5,  Ex.  5. 

"HO.    Show  that  the  equation  of  the  line  through  the  point 

(a?!,  y^)  parallel  to  the  line 

(3)  Ax  +  By  =  C 

is  Ax  +  By  =  Ax±  +  By±. 

Show  that  the  equation  of  the  line  through  the  point 
perpendicular  to  the  line  (3)  of  Ex.  10  is 
Bx  —  Ay  =  Bxy— 


Angle  between  Two  Lines.    Let  L±  and  L2 

lines,  whose  slopes  are,  respectively, 

A!  =  tan  #1,         and         \2  =  tan  62. 
To  find  the  angle,  <£,  from  L±  to  L2. 


Since 


THE   STRAIGHT   LINE 

y 


39 


j.        n         n 
9  =  "2  —  Vly 

it  follows  from  Trigonometry  that 


and  hence  that 


(1) 


tan  d>  =  — 


-X,     ^ 


FIG.  6 


1  -I-  AjA2 

The  angle  <f>  is  the  angle  from  LI  to  L<>.  That  is,  it  is  the 
angle  through  which  LI  must  be  rotated  in  the  positive  sense, 
about  the  point  A,  in  order  that  it  coincide  with  L».  In  par- 
ticular, we  agree  to  take  it  as  the  smallest  such  angle,  always 
less,  then,  than  180°  :  0  <  <f>  <  180°.* 

If  Ll  and  Z/2  are  perpendicular,  then,  by  (2),  §  6,  A2  =  —  V^-i 
and  1  +  A.^2  =  0.  Consequently,  cot  <£,  which  is  equal  to  the 
reciprocal  of  the  right-hand  side  of  (1),  has  the  value  zero,  and 
so  <j>  =  90°. 


Example.     Let  L±  and  L2  be  given  by  the  equations, 


A: 


Here  \i  =  2  and  A.2  =  —  3,  and  (1)  becomes 

3-2 


tan  <   = 


=  1. 


is  45° 


1-6 
Hence  the  angle  <£  from  LI  to 

In  deducing  (1)  it  was  assumed  that  L1  and 
Lz  both  have  slopes.  If  this  is  not  the  case, 
at  least  one  of  the  lines  is  parallel  to  the  axis 
of  y  and  no  formula  is  needed.  The  angle 
</>  may  be  found  directly.  Suppose,  for  ex- 
ample, that  LI  and  L2  are,  respectively, 

x  +  2  =  0        and        x  —  y  —  1. 

*  The  figure  shows  LI  and  L2  as  intersecting  lines,  but  formula  (1)  and 
the  deduction  of  it  are  valid  also  in  case  LI  and  L2  are  parallel.     In  this 


FIG.  7 


40  ANALYTIC   GEOMETRY 

Then  L^  is  parallel  to  the  axis  of  y,  and  Lz  is  inclined  at  an 
angle  of  45°  to  the  positive  axis  of  x,  since  Xj  =  1.  Conse- 
quently, <£  =  135°. 


In  each  of  the  following  exercises  determine  whether  the 
given  lines  are  mutually  parallel  or  perpendicular,  and  in  case 
they  are  neither,  find  the  angle  from  the  first  line  to  the  second. 

M.   z  +  2y=3,  x  +  2y=4. 

V   2a?  —     +  5  =  0,  4<c  -  2    -  7  =  0. 


6, 

7. 

8.   2o;-32/  = 


10.   2z- 


By  the  method  of  this  paragraph  determine  each  of  the 
three  angles  of  the  triangle  whose  sides  have  the  equations 

x  —  2y  —  6  =  Q,        2x-fy-4  =  0,        3x  —  y  +  3  =  Q. 
\  Check  your  results  by  adding  the  angles. 

T.2.   Prove  that  if  Lt  and  L2  are  represented  by  the  equations 
^  :  Ajx  +  By  +  d  =  0, 

Lz: 
then 


^^2  +  -01-02 

What  can  you  say  of  Lv  and  i2  if  ^1-82  —  -^2^1  =  0  ?    If 


case,  we  take  the  angle  from  LI  to  Lz  as  0°  —  ,  not  as  180°,  as  is  conceiv- 
able. Hence  arises  the  sign  <  (less  than  or  equal  to)  in  the  place  in 
which  it  stands  in  the  double  inequality. 


THE   STRAIGHT   LINE  41 

13.  Show  that  the  formula  of  Ex.  12  for  tan  <£  is  valid  even 
if  one  or  both  of  the  lines  has  no  slope,  i.e.  is  parallel  to  the 
axis  of  y. 

8.  Distance  of  a  Point  from  a  Line.  Let  P:(xlt  y^  be  a 
given  point  and  let 

L: 


be  a  given  line.     To  find  the  dis- 
tance, .D,  of  P  from  L. 

Drop  a  perpendicular  from  P 
on  the  axis  of  x,  and  denote  the 
point  in  which  it  cuts  L  by  Q. 
The  abscissa  of  Q  is  x±.  Denote 
its  ordinate  by  yq.  Then 


o 


FIG.  8 


Since  Q :  (xlf  y^)  lies  on  Z»,  its  coordinates  satisfy  the  equation 
of  L ;  thus 

Axl  +  ByQ+C=0. 

Solving  this  equation  for  y^,  we  find  : 


Hence 

(1)  QP  =  —  — • 

Let  6  be  the  slope-angle  of  L  and  form  the  product  QP  cos  0. 
One  or  both  of  the  factors  of  this  product  may  be  negative, 
according  to  the  positions  of  P  and  L.*  But'  always  the 
numerical  value  of  the  product  is  equal  to  the  distance  D : 

(2)  D=\QPco*0\. 

This  is  clear  in  case  P  and  L  are   situated  as   in   Fig.    8 ; 

*  There  are  four  essentially  different  positions  for  P  and  L,  for  L  may 
have  a  positive  or  a  negative  slope,  and  P  may  lie  on  the  one  or  on  the 
other  side  of  L. 


42 


ANALYTIC   GEOMETRY 


the  student  should  draw  the  other  typical  figures  and  show 
that  for  them,  also,  (2)  is  valid. 
Since  the  slope  of  L  is 


we  have 


A=tan0=-  — , 
B 


-s  sec1 6  =  1  -f  tan-  0  = 


B* 


~Z  cos  0  =  ± 


B 


4- 


It  is  immaterial  to  us  which  sign  in  (3)  is  the  proper  one. 
For,  according  to  (2),  we  have  now  to  multiply  together  the. 
values  of  QP  and  cos  0,  as  given  by  (1)  and  (3),  and  take  the 
numerical  value  of  the  product.  The  result  is  the  desired 
formula : 


(4) 

D  =  +  — 

V^2  + 

where,  in  the  second  formula,  that  sign  is  to  be  chosen  which 
makes  the  right-hand  side  positive. 

Example.     The  distance  of  the  point  (3,  —  2)  from  the  line 
is 


__ 

V32  +  42  V25 

The  deduction  of  formula  (4)  involves  division  by  B  and 
hence  tacitly  assumes  that  B  =j=  0,  i.e.  that  L  is  not  parallel  to 
the  axis  of  y.  The  formula  holds,  however,  even  when  L  is 
parallel  to  the  axis  of  y.  For,  in  this  case  it  is  clear  from  a 
figure  that 


and  (4)  reduces  precisely  to  this  when  B  =  0. 


THE   STRAIGHT   LINE 


43 


Ans.    2f. 
Ans.   2VIO,  or  6.32. 


EXERCISES 

In  each  of  the  first  seven  exercises  find  the  distance  of  the 
given  point  from  the  given  line. 

Point  Line 

*.   (5,2), 
2.    (2,3), 
V    (6,  -1), 
—  4.    (3,4), 

.   (-2, -5), 
.    Origin, 
.    Origin, 

.   Find  the  lengths  of  the  altitudes  of  the  triangle  with 
vertices  in  the  points  (2,  0),  (3,  5),  (—1,  2). 

9.   Area  of  a  Triangle.     Let  a  triangle  be  given  by  means  of 
its  vertices  (x^  y^,  fa,  y2),  fa,  yz\       y 
To  find  it's  area. 

Drop  a  perpendicular  from  one 
of  the  vertices,  as  (x3,  y3~),  on  the 
opposite  side.  Then  the  required 
area  is 


3x  +  5  =  0. 
2/  =  0. 

x  +  y  —  1  =  0. 


FIG.  9 


where  D  denotes   the   length   of 
the  perpendicular  and  E,  the  length  of  the  side  in  question. 
By  Ch.  I,  §  3,  we  have 

XT' 
JSj  = 


D  is  the  distance  of  (x3,  yz)  from  the  line  joining  (a^,  y^ 
and  (x2,  y2).  The  equation  of  this  line,  as  given  by  (I)  or  (I'), 
§  1,  may  be  put  into  the  form : 

(2/2  —  yi)»  -  («2  -  *i)y  - 
Consequently,  by  (4),  §  8,  we  find : 


44  ANALYTIC   GEOMETRY 

Thus 


A  =  ± 

The  result  may  be  written  more  symmetrically  in  either  of 
the  forms 

(1)  A  = 
or 

(2)  -4  =  ±  |[(yi  -  2/2)a3  +  (y,  - 

where  in  each  case  that  sign  is  to  be  chosen  which  makes  the 
right-hand  side  positive. 

EXERCISES 

Find  the  area  of   the   triangle  whose   vertices   are   in   the 
points 

1.    (1,2),  (-1,2),  (-2,1). 
/"2.    (5,3),  (-3,4),  (-2,-!). 

3.  (1,  2),  (2,  1),  (0,  0). 

Find  the  area  of  the  triangle  whose  sides  lie  along  the  lines 

4.  x  —  y  =  0,        x  +  y  =  0,        2x  +  y  —  3  =  Q. 

5.  2x  +  y-6  =  Q,        x-y  +  3  =  Q,         x-2y-8  =  0. 

6.  Find  the  area  of  the  convex  quadrilateral  whose  vertices 
are  in  the  points  (4,  2),  (-  1,  4),  (-  3,  -  2),  (5,  -  8). 

7.  What  do  formulas  (1)  and  (2)  become  when  one  of  the 
vertices,  say  (#3,  t/3),  is  in  the  origin  ? 

Ans.   A  = 


10.  General  Theory  of  Parallels  and  Perpendiculars.  Iden- 
tical Lines.*  The  line  through  the  point  (xlt  y^  parallel  to 
the  line 

(1)  Ax  +  By  =  C, 

has  the  equation,  according  to  §  6,  Ex.  10, 
Ax  +  By  =  Ax]_  +  By^ 

*  The  discussion  in  the  class-room  of  the  subjects  treated  in  this  and  the 
following  paragraph  may  well  be  postponed  until  the  need  for  them  arises. 


THE   STRAIGHT   LINE  45 

This  equation  is  of  the  form 

(2)  Ax  +  By  =  C', 

since  the  constant  Ax^  +  By-^  may  be  denoted  by  the  single 
letter  C". 

Conversely,  equations  (1)  and  (2),  for  C'  =J=  C,  always  repre- 
sent parallel  lines.  For,  if  -5=^0,  the  lines  have  the  same 
slope,  —  A/B  ;  if  B  =  0,  A  cannot  be  zero,  and  the  lines  are 
parallel  to  the  axis  of  y  and  hence  to  each  other. 

THEOREM  1.  Ttvo  lines  are  parallel  when  and  only  when  their 
equations  can  be  written  in  the  forms  (1)  and  (2),  where  C  =£  C". 

The  line  through  the  point  (x^  yi),  perpendicular  to  the 
line  (1),  has  the  equation  (§  6,  Ex.  11)  : 

Bx  —  Ay  =  BXi  —  Aylt 
and  this  equation  is  of  the  form 

(3)  Bx  —  Ay  =  C'. 

Let  the  student  show,  conversely,  that  equations  (1)  and  (3) 
always  represent  perpendicular  lines. 

THEOREM  2.  Two  lines  are  perpendicular  when  and  only 
when  their  equations  can  be  ivritten  in  the  forms  (1)  and  (3). 

The  equations  of  two  parallel  lines  can  always  be  written 
in  the  forms  (1)  and  (2).  But  they  need  not  be  so  written. 

Thus  the  lines, 

2x-y=-l, 

6x-3y  =  2, 

are  parallel,  though  the  equations  are  not  in  the  forms  (1)  and 
(2).     The  coefficients  of  the  terms  in  x  and  y  are  not  respec- 
tively equal.    They  are,  however,  proportional  :  2  :  6=  —  1  :  —  3. 
This  condition  holds  in  all  cases.     For  the  two  lines 


2  =  0, 
we  may  state  the  theorem  : 


46  ANALYTIC   GEOMETRY 

THEOREM  3.     Tlie  lines  LI  and  L2  are  parallel  *  if  and  only  if 

Al  :  A2  =  B!  :  B*. 

• 

For,  LI  and  L2  are  parallel  if  and  only  if  the  angle  <£  be- 
tween them,  as  defined  in  §  7,  is  zero  ;  but,  according  to  §  7, 
Ex.  12,  <f>,  or  better,  tan  <£,  is  zero,  when  and  only  when 
AiBz  —  A2Bi  =  0.  But  this  equation  is  equivalent  to  the  pro- 
portion AI  :  A2  =  Bl  :  B2. 

As  a  second  consequence  of  §  7,  Ex.  12,  we  obtain  the  fol- 
lowing theorem. 

THEOREM  4.  The  lines  LI  and  L2  are  perpendicular  if  and 
only  if 

AtA2  +  BiB2  =  0. 

Identical  Lines.  Two  equations  do  not  have  to  be  identically 
the  same  in  order  to  represent  the  same  line.  For  example, 
the  equations, 


represent  the  same  line.  The  corresponding  constants  in 
them  are  not  equa!4  but  they  are  proportional.  We  have, 
namely, 

2:6  =  -l:  -3  =  1:3, 

or,  what  amounts  to  the  same  thing, 

2:-l:l=6:  -3:3. 

This  condition  is  general.     We  formulate  it  as  a  theorem  : 
THEOREM  5.     The  lines  LI  and  L2  are  identical  if  and  only  if 

AI  '.  A2^Bi  '.  B%  =  GI  :  ^2, 
or  A1:B1:C1  =  A2  :  B2  :  C2. 

For,  LI  and  L2  are  the  same  line  when  and  only  when  they 
have  the  same  slope  and  the  same  intercept  on  the  axis  of  y, 
that  is,  when  and  only  when 

_^i  =  _4*         and         _^L  =  _^?, 
BI         .Bo  BI         -B2 

*  Or,  in  a  single  case,  identical.     Cf  .  Th.  5. 


THE   STRAIGHT   LINE  47 

or  AI  :  A2  =  Bl:  B2          and       B1  :  B2=Ci:  C2) 

or,  finally,  Al  :  Az  =  B^  :  B2  =  Ci  :  C2. 

This  proof  assumes  that  Uj  =£  0  and  JB2  =£  0.  The  proof, 
when  this  is  not  the  case,  is  left  to  the  student. 

EXERCISES 

1.  Prove  Th.  3  directly,  without  recourse  to  the  results 
of  §7. 

2.  The  same  for  Th.  4. 

See  also  Exs.  15,  16,  17,  18  at  the  end  of  the  chapter. 

11.   Second  Method  of  Finding  Parallels  and  Perpendiculars. 

Problem  1.  To  find  the  equation  of  a  line  parallel  to  the 
given  line 

(1)  Ax  +  By=C, 

and  satisfying  a  further  condition. 

By  §  10,  Th.  1,  the  desired  equation  can  be  written  in  the 
form 

(2)  Ax  +  By=C', 

where  C'  is  to  be  determined  by  the  further  condition. 

Example.  Consider  the  first  example  treated  in  §  6.  In 
this  case  the  equation  of  the  desired  line  can  be  written  in  the 
form 

3x-2y  =  k, 

where  we  have  replaced  the  C'  of  (2)  by  "k.  The  "further 
condition,"  by  means  of  which  the  value  of  k  is  to  be  deter- 
mined, is  that  the  line  go  through  the  point  (1,  2),  Hence 
x  =  i,  y  =  2  must  satisfy  the  equation  of  the  line,  or 

3  •  1  -  2  •  2  =  Jc. 
Consequently,  k  =  —  1,  and  the  equation  of  the  line  is 


Problem  2.     To  find  the  equation  of  a  line  perpendicular  to 
the  given  line  (1)  and  satisfying  a  further  condition. 


48  ANALYTIC   GEOMETRY 

By  §  10,  Th.  2,  the  desired  equation  can  be  written  in  the 
form 

(3)  Bx-Ay=  C", 

where  C"  is  to  be  determined  by  applying  the  further  condition. 
This  condition  does  not  always  have  to  be  that  the  line 
should  go  through  a  given  point.  It  may  be  any  single  con- 
dition, not  affecting  the  slope  of  the  line,  which  it  seems  de- 
sirable to  apply.  We  give  an  example  illustrating  the  method 
in  such  a  case. 

Example.     To  find  the  equation  of  the  line  perpendicular  to 
2s -y -4  =  0 

and  cutting  from  the  first  quadrant  a  triangle  whose  area  is  16. 
Equation  (3)  may,  in  this  case,  be  written  as 

(4)  x  +  2  y  =  k. 

We  are  to  determine  k  so  that  the  line  (4)  cuts  from  the  first 
quadrant  a  triangle  of  area  16.  The  intercepts  of  the  line  (4) 
are  k  and  ^k,  and  hence  the  area  of  the  triangle  in  question  is 
^kz.  Accordingly,  i&2  =  16,  and  k  =  ±8.  But  the  line  cuts 
the  first  quadrant  only  if  k  is  positive,  and  so  we  must  have 
7;  =  8.  The  equation  of  the  desired  line  is,  then, 


EXERCISES 

1.   Work  Exs.  1-4,  8,  9  of  §  6  by  this  method. 
r  2.    There  are  two  lines  parallel  to  the  line 

x  —  2y  =  6 

and   forming  with   the   coordinate   axes   triangles  of  area  9. 
Find  their  equations. 

/  \SJ  Find  the  equations  of  the  lines  parallel  to  the  line  of 
Ex.  2  and  3  units  distant  from  it. 

Suggestion.  Write  the  equation  of  the  required  line  in  the 
form  (2)  and  demand  that  the  distance  from  it  of  a  chosen 
point  of  the  given  line  be  3. 


THE   STRAIGHT   LINE  49 

4.    Find  the  equations  of  the  lines  parallel  to  the  line 


and  2  units  distant  from  the  origin. 

J/  5.    The  same  as  Ex.  2,  if  the  lines  are  to  be  perpendicular, 
instead  of  parallel,  to  the  given  line. 

6.  The  same  as  Ex.  4,  if  the  lines  are  to  be  perpendicular, 
instead  of  parallel,  to  the  given  line. 

7.  A  line  is  parallel  to  the  line  3x-\-  2y  —  6  =  0,  and  forms 
.a  triangle  in  the  first  quadrant  with  the  lines, 

x  —  2  y  —  0        and         2  x  —  y  =  0, 

whose  area  is  21.     Find  the  equation  of  the  line.  ' 

Ans.   3z  +  2y-28=0. 


1.  Find  the  equation  of  theWine  whose  intercepts  are  twice 
those  of  the  line  2x  —  3y  —  6  =  0. 

2.  Find  the  equation  of  the  line  having  the  same  intercept 
on  the  axis  of  x  as  the  line  V3#  —  y  —  3  =  0,  but  making  with 
that  axis  half  the  angle. 

3.  Find  the  equation  of  the  line  joining  the  point  (7,  —  2) 
with  that  point  of  the  line  2x  —  y  =  S  whose  ordinate  is  2. 

4.  A  perpendicular  from  the  origin  meets  a  line  in  the  point 
(5,  2).     What  is  the  equation  of  the  line? 

r  5.  The  coordinates  of  the  foot  of  the  perpendicular  dropped 
from  the  origin  on  a  line  are  (a,  6).  Show  that  the  equation 
of  the  line  is 

ax  +  by  =  a2  -f-  ft2. 

6.  The  line  through  the  point  (5,  —3)  perpendicular  to  a 
given  line  meets  it  in  the  point  (—3,  2).     Find  the  equation 
of  the  given  line. 

7.  Prove  that  the  line  with  intercepts  6  and  3  is  perpen- 
dicular to  the  line  with  intercepts  3  and   —  6.     Is  it  also  per- 
pendicular to  the  line  with  intercepts  —  3  and  6  ? 


50  ANALYTIC   GEOMETRY 


.   Prove  that  the  line  with  intercepts  a  and  6  is  perpen- 
dicular to  the  line  with  intercepts  b  and  —  a. 

9.    Show  that  the  two  points  (5,  2)  and  (6,  —  15)  subtend  a 
right  angle  at  the  origin. 

I/    10.   Prove  that  the  two  points,  (xl}  y^)  and  (a^,  y2),  subtend  a 
right  angle  at  the  origin  when,  and  only  when,  x^  +  y\yz  =  0. 

11.  Do  the  points  (6,  •—  1)  and  (—3,4)  subtend   a  right 
angle  at  the  point  (4,  6)  ?    At  the  point  (—  4,  —  2)  ? 

12.  Given  the  triangle  whose  sides  lie  along  the  lines, 
x  —  2y  +  6  =  Q,        2x-y  =  3,        x  +  y  —  3  =  0. 

Fin(jl  the  coordinates  of  the  vertices  and  the  equations  of  the 
lines  through  the  vertices  parallel  to  the  opposite  sides. 
tX   13.   Two  sides  of  a  parallelogram  lie  along  the  lines, 


A  vertex  is  at  the  point  (—2,  1).     Find  the  equations  of  the 
other  two  sides  (produced). 

14.    One  side  of  a  rectangle  lies  along  the  line, 


A  vertex  on  this  side  is  at  the  point  (1,  1)  and  a  second  vertex 
is  at  (2,  —  1).  Find  the  equations  of  the  other  three  sides 
(ppgduced). 

l5J   For  what  value  of  A  will  the  two  lines, 


'(a)  be  parallel?     (6)  be  perpendicular  ? 
\/   16.   For  what  value,  or  values,  of  m  will  the  two  lines, 

4a;  —  my  +  6  =  0,         x  +  my  +  3  =  0, 
(a)  be  parallel  ?     (6)  be  perpendicular  ? 

17.   For  what  value  of  m  will  the  two  equations, 
mx  +  y  +  5  =  0,        4a;  +  my  +  10  =  0, 
represent  the  same  line  ? 


THE   STRAIGHT   LINE  51 

'18.   For  what  pairs  of  values   for  k  and   I   will   the   two 
equations, 


represent  the  same  line  ? 

19.  The  equations  of  the  sides  of  a  convex  quadrilateral  are 

aj  =  2,        y  =  4,        y  =  x,        2y  =  x. 

Find  the  coordinates  of  the  vertices  and  the  equations  of  the 
diagonals. 

20.  Find  the  equation  of   the  line   through   the   point   of 
intersection  of  the  lines, 

3x  -  5y  —  11  =  0,         2x—7y  =  ll, 
and  having  the  intercept  —  5  on  the  axis  of  y. 

21  )  Find  the  equation  of  .the  line  through  the  point  of  in- 
fection of  the  lines, 


and  perpendicular  to  the  first  of  these  two  lines. 

22.    Find  the  distance  between  the  two  parallel  lines, 


Suggestion.     Find  the  distance  of  a  chosen  point  of  the  first 
line  from  the  second. 
ir     23.   Let 

Ax  +  By  +  C  =  0         and         Ax  +  By  -f  C"  =  0 

be  any  two  parallel  lines.     Show  that  the  distance  between 
them  is 

\G'-C\  or  ±     C'-C 

' 


24.  There  are  two  points  on  the  axis  of  x  which  are  at  the 
distance  4  from  the   line   2x  —  3y  —  4  =  0.     What  are  their 
coordinates  ? 

25.  Find  the  coordinates  of  the  point  on  the  axis  of  y  which 
is  equidistant  from  the  two  points  (3,  8),  (—2,  5). 


52  ANALYTIC   GEOMETRY 


There  are  two  lines  through  the  point  (1,  1),  each 
cutting  from  the  first  quadrant  a  triangle  whose  area  is  1\. 
Find  their  slopes.  Ans.  —  ^,  —  2. 

27.  Find  the  equation  of  the  line  through  the  point  (3,  7) 
such  that  this  point  bisects  the  portion  of  the  line  between 
the  axes.  Ans.   Ix  +  3y  —  42  =  0. 

28.  The  origin  lies  on  a  certain  line  and  is  the  mid-point  of 
that  portion  of  the  line  intercepted  between  the  two  lines, 

3x  —  5y  =  6,        4a  +  2/  +  6  =  0. 

Find  the  equation  of  the  line.  Ans.   x  +  Qy  =  0. 

29.  The  line 

(1)  3x  -  8y  +  5  =  0 

goes  through  the  point  (1,  1).     Find  the  equation  of  the  line 

(2)  through  this  same  point,  if  the  angle  from  the  line  (1)  to 
the  line  (2)  is  45°.  Ans.   llx  —  5y  -  6  =  0. 

30.  Find  the  equations  of  the  two  lines  through  the  origin 
making  with  the  line  2x  —  3y  =  0  angles  of  60°. 


CHAPTER   III 
APPLICATIONS 

1.  Certain  General  Methods.  Lines  through  a  Point.  In 
many  theorems  and  problems  of  Plane  Geometry  the  question 
is  to  show  that  three  lines  pass  through  a  point.  Plane  Geom- 
etry affords,  however,  no  general  method  for  dealing  with  this 
question.  Each  new  problem  must  be  discussed  as  if  it  were 
the  first  of  this  class  to  be  considered. 

Analytic  Geometry,  on  the  other  hand,  affords  a  universal 
method,  whereby  in  any  given  case  the  question  can  be  settled. 
For,  from  the  data  of  the  problem,  the  equation  of  each  of  the 
lines  can  be  found.  These  will  all  be  linear,  and  can  be  writ- 
ten in  the  form 

^  :  Ajx  +  B$  +  d  =  0, 

L2  :  A.2x  +  Bzy  +  <72  =  0, , 

L3 :  Asx  +  B3y  +  C3  =  0- 

The  coordinates  of  the  point  of  intersection  of  two  of  these 
lines,  as  Lt  and  L2)  can  be  found  by  solving  the  corresponding 
equations,  regarded  as  simultaneous,  for  the  unknown  quanti- 
ties x  and  y.  Let  the  solution  be  written  as 

x  =  x',  y  =  y'. 

The  third  line,  L3)  will  pass  through  this  point  (a/,  ?/'),  if  and 
only  if  the  coordinates  of  the  latter  satisfy  the  equation  of  L3; 
i.e.  if  and  only  if 

A3x'  +  B3y'  +  C3  =  0. 

Points  on  a  Line.  A  second  question  which  presents  itself 
in  problems  of  Plane  Geometry  is  to  determine  when  three 

53 


54 


ANALYTIC   GEOMETRY 


points  lie  on  a  straight  line.  Here,  again,  the  reply  of  Analytic 
Geometry  is  methodical  and  universal.  From  the  data  of  the 
problem  it  will  be  possible  in  any  given  case  to  obtain  the 
coordinates  of  the  three  points.  Call  them 


Now,  we  know  how  to  write  down  the  equation  of  a  line 
through  two  of  them,  as  (xl9  y^  and  (a^,  y2~).  This  equation 
will  always  be  linear,  and  can  be  written  in  the  form 

Ax  +  By  +  C  =  0. 

The  third  point,  (x3 ,  ys~),  will  lie  on  this  line  if  and  only  if  its 
coordinates  satisfy  the  equation  of  the  line;  i.e.  if  and  only  if 

The  student  should  test  his  understanding  of  the  foregoing 
theory  by  working  Exs.  1-6  at  the  end  of  the  chapter. 

2.  The  Medians  of  a  Triangle.  We  recall  the  proposition 
from  Plane  Geometry,  that  the  medidns  of  a  triangle  meet  in  a 
point.  The  proof  there 
given  is  simple,  provided 
one  remembers  the  con- 
struction lines  it  is  desir- 
able to  draw.  By  means, 
however,  of  Analytic  Ge- 
ometry we  can  establish  j?'(-i,3) 
the  proposition,  not  by 
artifices,  but  by  the  natural 
and  direct  application  of 
the  general  principle  enun- 


CV(0,6) 


A:(2,3) 


£.-(4,0) 


FIG.  1 


ciated     in    the     preceding 
paragraph. 

The  first  step  consists  in 
the  choice  of  the  coordinate  axes.  This  choice  is  wholly  in 
our  hands,  and  we  make  it  in  such  a  way  as  to  simplify  the 
coordinates  of  the  given  points.  Thus,  clearly,  it  will  be  well 


APPLICATIONS  55 

to  take  one  of  the  axes  along  a  side  of  the  triangle.  Let  this 
be  the  axis  of  x. 

A  good  choice  for  the  axis  of  y  will  be  one  in  which  this 
axis  passes  through  a  vertex.  Let  this  be  the  vertex  not  on 
the  axis  of  x. 

We  begin  with  a  numerical  case,  choosing  the  vertices  A,  B, 
C  at  the  points  indicated  in  the  figure. 

The  Equations  of  the  Medians.  Consider  the  median  AA'. 
One  point  on  this  line  is  given,  namely  A  :  (—  2,  0).  A  second 
point  is  the  mid-point  A'  of  the  line-segment  BC.  By  Ch.  I, 
§  5,  the  coordinates  of  A'  are  (2,  3). 

The  student  can  now  solve  for  himself  the  problem  of  finding 
the  equation  of  the  line  L}  through  A  :  (—  2,  0)  and  A'  :  (2,  3). 
The  answer  is, 
A:  3x-4y  +  6  =  0. 

In  a  precisely  similar  way  the  coordinates  of  B'  are  found 
to  be  (  —  1,  3),  and  the  equation  of  the  median  BB'  is 


Finally,  the  coordinates  of  C1  are  (1,  0),  and  the  equation  of 
the  median  CO'  is 


The  Point  of  Intersection  of  the  Medians.  The  next  step  con- 
sists in  finding  the  point  in  which  two  of  the  medians,  as  LI 
and  L2,  intersect.  The  coordinates  of  this  point  will  be  given 
by  solving  as  simultaneous  the  equations  of  these  lines  : 

*  3x  -  4y  +  6  =  0, 


The  solution  is  found  to  be  : 


And  now  the  third  median,  L3,  will  go  through  this  point, 
(|-,  2),  if  the  coordinates  of  the  point  satisfy  the  equation  of  L3, 


56 


ANALYTIC-  GEOMETRY 


On  substituting  for  x  in  this  equation  the  value  ^  and  for  y 
the  value  2,  we  are  led  to  the  equation 


This  is  a  true  equation,  and  hence  the  three  lines  LI,  L2,  and 
L3  pass  through  the  same  point. 

Remark.  It  can  be  shown  by  the  formulas  of  Ch.  I,  §  6, 
that  the  above  point  (-|,  2)  trisects  each  of  the'  medians  A  A', 
BB',  and  CO1. 

\.s\  EXERCISES 

if     •   1 

IV    nJ  Taking  the  same  triangle  as  before,  choose  the  axis  of  x 

along  the  side  AB,  but  take  the  axis  of  y  through  A.     The 
coordinates  of  the  vertices  will  then  be  : 

,,        -4:  (0,0);         B  :  (6,  0)  ;         C:(2,6). 

Prove  the  theorem  for  this  triangle. 

2.  The  vertices  of  a  triangle  lie  at  the  points  (0,  0),  (3,  0), 
(0,  9).     Prove  that  the  medians  meet  in  a  point. 

3.  Continuation.     The   General   Case.     We  now  proceed  to 
prove  the  theorem  of  the  medians  for  any  triangle,  ABC.     Let 


the  axes  be  chosen  as  in  the  text  of  §  2.     Then  the  coordinates 
of  A  will  be  (a,  0),  where  a  may  be   any  number  whatever, 


APPLICATIONS  57 

positive,  negative,  or  zero.  The  coordinates  of  B  will  be  (6,  0), 
where  b  may  be  any  number  distinct  from  a : 

b  =£  a,         or         a  —  b-=f=.  0.* 

Finally,  the  coordinates  of  C  can  be  written  as  (0,  c),  where  c 
is  any  positive  number. 

Next,  find  the  coordinates  of  A',  B1,  C'.  They  are  as  shown 
in  the  figure. 

The  equation  of  LI  is  given  by  Ch.  II,  (I),  where 

(aa,  &)  =  (<»,  0);         (z2,  3fe) 

#—  a_T/  —  0 

!*«:  6~~      ~c      I' 

—  a     —  u 

2  2 

or 

Z/i :  cic  +  (2  a  —  b}y  =  ac. 

The  equation  of  L2  can  be  worked  out  in  a  similar  manner. 
But  it  is  not  necessary  to  repeat  the  steps,  since  interchanging 
the  letters  a  and  b  interchanges  the  points  A  and  B,  and  also 
A'  and  B'.  Thus  L^  passes  over  into  L2.  Hence  the  equation 
of  Lz  is : 

L2 :  ex  +  (2  b  —  a)y  =  be. 

The  line  L3  is  determined  by  its  intercepts,  £(a  +  6)  and  c ; 
by  Ch.  II,  (IV),  its  equation  is  found  to  be : 

L3 :  2  ex  +  (a  -+-  V)y  =  (a  +  6)c. 

To  find  the  coordinates  of  the  point  in  which  LI  and  L2  in- 
tersect, solve  as  simultaneous  the  equations  of  L±  and  L2 : 

\cx+(2a  —  b)y  =  ac, 
[  ex  +  (2  6  —  a)y  =  be. 
The  result  is  : 

a  +  b  c 

x  =  — ! — ,  y  =-• 

3  3 

*  The  figure  has  been  drawn  for  the  case  in  which  a  is  negative  and  b 
positive. 


58  ANALYTIC   GEOMETRY 

Finally,  to  show  that  this  point,  (  a        ,  -\  lies  on  LZ)  sub- 

\    3        3J 

stitute  its  coordinates  in  the  equation  of  L3  : 


Since  this  is  a  true  equation,  the  point  lies  on  the  line,  and  we 
have  proved  the  theorem  that  the  medians  of  a  triangle  pass 
through  a  point. 

That  this  point  trisects  each  median  can  be  proved  as  in  the 
special  case  of  the  preceding  paragraph,  by  means  of  Ch.  I,  §  6. 
The  details  are  left  to  the  student. 

EXERCISE 

Prove  the  theorem  of  the  medians  by  taking  the  coordinate 
axes  as  in  the  first  exercise  of  the  preceding  paragraph.  Here, 
the  vertices  are 

^:(0,0);         £:(a,0);          C  :  (6,  c), 

where  a  may  be  any  number  not  0,  6  any  number  whatever, 
and  c  any  positive  number.  Draw  the  figure,  and  write  in  the 
coordinates  of  each  point  used. 

4.  The  Altitudes  of  a  Triangle.  Another  proposition  of 
Plane  Geometry  is,  that  the  perpendiculars  dropped  from  the 
vertices  of  a  triangle  on  the  opposite  sides  meet  in  a  point. 

The  proof  of  the  proposition  by  Analytic  Geometry  is  direct 
and  simple.  Let  us  begin  with  a  numerical  case,  taking  the 
triangle  of  Fig.  1.  One  of  the  perpendiculars  is,  then,  the 
axis  of  y,  and  so  all  that  is  necessary  to  show  is  that  the  other 
two  meet  on  this  axis,  or  that  the  ^coordinate  of  their  point 
of  intersection  is  0. 

The  equation  of  the  line  BC  can  be  written  down  at  once  in 
terms  of  its  intercepts  : 

f  +  |  =  l,         or 
4     6 


APPLICATIONS  59 

The  slope  of  this  line  is  A  =  —  f .  The  slope  of  any  line 
perpendicular  to  it  is  A'  =  f .  Hence  the  equation  of  LI,  the 
perpendicular  which  passes  through  the  point  A  :  (—  2,  0),  is 


or 

L!  :  2x  —  3y  +  4  =  0. 

In  a  similar  manner  the  student  can  obtain  the  equation  of 
the  perpendicular  L2  from  B  on  the  side  AC.  It  is, 

Z/2 :  x  -{-3y  —  4  =  0. 

On  computing  the  avcoordinate  of  the  point  in  which  LI  and 
L2  intersect,  it  is  found  that  x  =  0,  and  hence  the  proposition 
is  established  for  this  triangle. 

Remark.  For  use  in  a  later  problem  it  is  necessary  to 
know  the  exact  point  in  which  the  perpendiculars  meet.  It  is 
readily  shown  that  this  point  is  (0,  -|). 

EXERCISES 

1.  Prove  the  above  proposition  for  the  special  triangle  con- 
sidered, choosing  the  coordinate  axes  as  in  Ex.  1  of  §  2. 

2.  Prove  the  proposition  for  the  triangle  of  Ex.  2,  §  2. 

3.  Prove  the  proposition  for  the  general  case,  choosing  the 
axes  as  in  Fig.  2.     First  show  that  the  equation  oi  the  perpen- 
dicular LI  from  A  on  BC  is 

L!  :  bx  —  cy  =  ab, 

and  that  the  equation  of  the  perpendicular  L>2  from  B  on  AC  is 

Lz :  ax  —  cy  =  ab. 

Then  show  that  these  lines  intersect  each  other  on  the  axis 

off. 

^  /p  Show  that  the  point  in  which  the  perpendiculars  in  the 

preceding  question  meet  is  (0,  —  —  }• 

V  CJ 

5.  Prove  the  theorem  of  the  altitudes,  when  the  axes  of 
coordinates  are  taken  as  in  the  exercise  of  §  3. 


60  ANALYTIC   GEOMETRY 

5.  The  Perpendicular  Bisectors  of  the  Sides  of  a  Triangle. 
It  is  shown  in  Plane  Geometry  that  these  lines  meet  in  a 
point.  Since  the  student  is  now  in  full  possession  of  the 
method  employed  in  Analytic  Geometry  for  the  proof  of  this 
theorem,  he  will  find  it  altogether  possible  to  work  out  that 
proof  without  further  suggestion.  Let  him  begin  with  the 
special  triangle  of  Fig.  1.  He  will  find  that  the  equations  of 
the  perpendicular  bisectors  of  the  sides  are  the  following  : 


L2: 

L3  :  x  -  1  =  0. 

These  lines  are  then  shown  to  meet  in  the  point  (1,  £). 

He  can  work  further  special  examples  corresponding  to  the 
exercises  at  the  end  of  §  2  if  this  seems  desirable. 

Finally,  let  him  work  out  the  proof  for  the  general  case,  tak- 
ing the  coordinate  axes  as  in  Fig.  2.  The  three  lines  will  be 
found  to  have  the  equations 

L!  :  bx-cy  =  |(62  -  c2), 

Z/2  :  ax  —  cy  =  |(a2  —  c2), 

L3  :  x  ==  |(a  +  6). 

They  meet  in  the  point 

fa  +  b     ab  4-  c2\ 
,     (   2  2c    )' 

EXERCISE 

Give  the  proof  when  the  axes  of  coordinates  are  taken  as  in 
the  exercise  of  §  3. 

6.  Three  Points  on  a  Line.  The  foregoing  three  propositions 
about  triangles  have  led  to  three  points,  namely,  the  three 
points  of  intersection  of  the  three  lines  in  the  various  cases. 
In  the  case  of  the  special  triangle  of  Fig.  1,  these  points  are 

(|,2);        "(0,1);         (I,*)- 


APPLICATIONS  61 

These  points  lie  on  a  straight  line.  Let  the  student  try  to 
prove  this  theorem  by  Plane  Geometry. 

The  proof  by  Analytic  Geometry  is  given  immediately  as  a 
direct  application  of  the  second  of  the  general  principles  enun- 
ciated in  the  opening  paragraph  of  the  chapter. 

Write  down  the  equation  of  the  line  through  two  of  these 
points,  —  say,  through  the  first  and  third.  It  is  found  to  be  : 

3x  -3y  +  4  =  0. 
The  coordinates  of  the  second  point, 

x  =  0,  y  =  |, 

are  seen  to  satisfy  this  equation,  and  the  proposition  is  proved. 

EXERCISES 

1.  Prove  the  proposition  for  the  general  case  (Fig.  2).  The 
points  have  been  found  to  be  : 

b    ab  -+-  c- 


fa  4-  b     c\      f~    _«& 
(3     '  3j'     V       "T 


2.  On  plotting  the  three  points  obtained  in  the  special  case 
discussed  in  the  text  it  is  observed  that  the  line-segment  de- 
termined by  the  extreme  points  is  divided  by  the  intermedi- 
ate point  in  the  ratio  of  1:2.  Prove  this  analytically.  Is  it 
true  in  general? 

EXERCISES  ON   CHAPTER   III 

Prove  that  the  three  lines, 

x  -  3y  —  5  =  0,     3x  +  4y  —  16  =  0,     4#  —  23y  +  7  =  0, 
go  through  a  point. 

2.   Prove  that  the  three  lines, 

ax  +  by  =  1,     bx  +  ay  =  1,     x  —  y  =  0, 

gaestiirough  a  point. 

f  5y  Prove1  that   the   three   points   (4,  1),    (—  1,    —  9),   and 

(27  —  3)  lie  on  a  line. 


62  ANALYTIC   GEOMETRY 

4.  Prove   that   the  three   points    (a,  6),  (b,  a),   and 
(—  a,  2a  -4-  &)  lie  on  a  line. 

5.  Find  the  condition  that  the  three  lines, 

bx  +  ay  =  2ab,     ax  +  by  =  a?  +  W,     Bx  —  2y  =  0, 
where  a2  is  not  equal  to  62,  meet  in  a  point. 

6.  Find  the  condition  that  the  three  points  (a,  6),  (b,  a), 
and  (2  a,  —  6),  where  a  is  not  equal  to  6,  lie  on  a  line. 

LINES  THROUGH  A  POINT 

7.  Show  that  the  line  drawn  through  the  mid-points  of  the 
parallel  sides  of  a  trapezoid  passes  through  the  point  of  inter- 
section of  the  non-parallel  sides. 

8.  Show  that,  in  a  trapezoid,  the  diagonals-  and  the  line 
.drawn  through  the  mid-points  of  the  parallel  sides  meet  in  a 
point. 

9.  A  right  triangle  has  its  vertices  A,  B,  and  0  in  the  points 
(4, 0),  (0,  3),  and  (0,  0).    The  points  A' :  (4,  -  4)  and  B' :  (-  3, 
3)  are  marked.     Prove  that  the  lines  AB',  BA',  and  the  per- 
pendicular from  0  on  the  hypothenuse  meet  in  a  point. 

10.  (Generalization  of  Ex.  9.)     Given  a  right  triangle  ABO 
with  the  right  angle  at  0.     On  the  perpendicular  to  OA  in 
the  point  A  measure  off  the  distance  AA',  equal  to  OA,  in  the 
direction  away  from  the  hypothenuse.     In  a  similar  fashion 
mark  the  point  B'  on  the  perpendicular  to  OB  in  B,  so  that 
BB'  =  OB.     Prove  that  the  lines  AB',  BA',  and  the  perpen- 
dicular from  0  on  the  hypothenuse  meet  in  a  point. 

11.  Let  P  be  any  point  (a,  a)  of  the  line  x  —  y  =  0,  other 
than  the  origin.     Through  P  draw  two  lines,  of  arbitrary  slopes 
AI  and  A2,  intersecting  the  cc-axis  in  A]  and  A2  and  the  y-axis 
in  Bl  and  B2  respectively.     Prove  that  the  lines  AiB.2  and 
A^Bi  will,  in  general,  meet  on  the  line  x  +  y  =  0. 

12.  If  on  the  three  sides  of  a  triangle  as  diagonals  paral- 
lelograms, having  their  sides  parallel  to  two  given  lines,  are 


APPLICATIONS  63 

described,  the  other  diagonals  of  the  parallelograms  meet  in  a 
point. 

Prove  this  theorem,  when  the  given  lines  are  the  coordinate 
axes,  and  the  triangle  has  as  its  vertices  the  points  (1,  6), 
(4,  11),  (9,  3). 

13.  Prove  the  theorem  of  the  preceding  exercise,  when  the 
given  lines  are  the  axes,  and  the  triangle  has  its  vertices  in  the 
points  (0,  0),  (a,  a),  (6,  c). 

POINTS  ON  A  LINE 

14.  Show  that  in  the  parallelogram  ABCD  the  vertex  D, 
the  mid-point  of  the  side  AB,  and  a  point  of  trisection  of  the 
diagonal  AC  lie  on  a  line. 

15.  Prove  that  the  feet  of   the   perpendiculars   from   the 
point  (2,  —  1)  on  the  sides  of  the  triangle  with  vertices  in  the 
points  (0,  0),  (3,  0),  and  (0,  1)  lie  on  a  line. 

16.  Prove  that  the  feet  of  the  perpendiculars  from  the  point 
(—1,4)  on  the  sides  of  the  triangle  with  vertices  in  the  points 
(2,  0),  (-  3,  0),  and  (0,  4)  lie  on  a  line. 

•   17.   Show  that  the  feet  of  the  perpendiculars  from  the  point 
( 0,  —  J  on  the  sides  of  the  triangle  with  vertices  in  the  points 

(a,  0),  (6,  0),  and  (0,  c)  lie  on  a  line. 

18.  Let  J/be  the  point  of  intersection  of  two  opposite  sides 
of  a  quadrilateral,  and  N,  the  point  of  intersection  of  the  other 
two  sides.     The  mid-point  of  MN  and  the  mid-points  of  the 
diagonals  lie  on  a  right  line. 

Prove  this  proposition  for  the  special  case  that  the  vertices 
of  the  quadrilateral  are  situated  at  the  points  (0,  0),  (8,  0),  (6, 4), 
(1,  6). 

19.  Prove  the  proposition  of  Ex.  18  for  the  general  case. 
Suggestion.     Take   the   axis  of  x  through  M  and  N,  the 

origin  being  at  the  mid-point.     The  equations  of  the  sides  can 
then  be  written  in  the  form 


64  ANALYTIC   GEOMETRY 

y  =  \i(x  -K),        y  =  As(aj  -  h\ 
y  =  A3(a  +  K),        y  =  A.4(>  +  h). 

20.  Let  0  be  the  foot  of  the  altitude  from  the  vertex  C  of 
the  triangle  ABC  on  the  side  AB.     Then  the  feet  of  the  per- 
pendiculars from  O  on  the  sides  BC  and  AC  and  on  the  other 
two  altitudes  lie  on  a  line. 

Prove  this  theorem  for  the  triangle  ABC  with  vertices  in 
the  points  (1,  0),  (-  4,  0),  (0,  2). 

21.  Prove  the  theorem  of  the  preceding  exercise   for   the 
triangle  with  vertices  in  the  points  (a,  0),  (6,  0),  (0,  c).     It 
will  be  found  that 

bci  62C    \ 


a2  +  c2'    a2  +  c2/    \6'J  +  c2'    &2  -j-  c2/ 
aft2         —  a6c\ 


are  the  coordinates  of  the  four  points  which  are  to  lie  on  a 
line,  and  that 

c(a  +  b~)x  -\-  (ab  —  c2~)y  =  abc 

is  the  equation  of  the  line. 


CHAPTER   IV 


THE  CIRCLE 

1.   Equation   of  the   Circle.     According  to  Ch.  I,  §  7,  the 

equation  of  the  circle  whose  center  is  at  the  origin,  and  whose 
radius  is  p,  is 

In  a  precisely,  similar  manner,  the  equation  of  a  circle  with 
its    center    at    an    arbitrary    point 
C:  (a,  ft)    of   the   plane,  the   length 
of  the  radius  being  denoted  by  p,  is 
found  to  be  : 

Example.     Find    the    equation   of    75 
the   circle   whose    center    is    at    the  FIG.  l 

point  (—  f,  0),  and  whose  radius  is  |. 

Here,  a  =  —  f ,  ft  =  0,  and  p  =  f .     Hence,  from  (2) : 


This  equation  can  be  simplified  as  follows : 

or,  finally, 

3x2  +  3t/2  +  8x  +  4  =  0. 


EXERCISES 

Find  the  equations  of  the  following  circles,  and  reduce  the 
results  to  their  simplest  form.     Draw  the  figure  each  time. 

1.  Center  at  (4,  6) ;  radius,  3. 

AM.   x2  +  y2  -  8a  -  12 y  +  43  =  0. 

2.  Center  at  (0,  —  2)  ;  radius,  2.         Ans.   a;2  +  yz  +  4y  =  0. 

65 


66  ANALYTIC   GEOMETRY 

3.  Center  at  (—  3,  0)  ;  radius,  3. 

4.  Center  at  (2,  —  4)  ;  radius,  8. 

5.  Center  at  (0,  f)  ;  radius,  f  . 

6.  Center  at  (3,  —  4)  ;  radius,  5. 

7.  Center  at  (—  5,  12)  ;  radius,  13. 

8.  Center  at  (•£,  —  f  )  ;  radius,  2. 

9.  Center  at  (—  f,  f)  ;  radius,  JJ-. 

10.  Center  at  (a,  0)  ;  radius,  a. 

11.  Center  at  (0,  a)  ;  radius,  a. 

12.  Center  at  (a,  a)  ;  radius,  a  V2. 

2.   A  Second  Form  of  the  Equation.     Equation  (2)  of  §  1  can 
be  expanded  as  follows  : 

x"-  +  y1  -  2ax  -  2py  +  a2  -|-  /S2  -  p2  =  0. 
This  equation  is  of  the  form 

(1)  tf  +  yt  +  Ax  +  By+C=Q. 

Let  us  see  whether,  conversely,  equation  (1)  always  repre- 
sents a  circle. 

Example   1.      Determine    the    curve    represented    by    the 
equation 

(2)  a;2  +  y*  +  2x-  6^  +  6  =  0. 
We  can  rewrite  this  equation  as  follows  : 

(Xs  +  2x        )  +  <y-67/        )  =  -6. 

The  first  parenthesis  becomes  a  perfect  square  if  1  is  added  ; 
the  second,  if  9  is  added.  To  keep  the  equation  true,  these 
numbers  must  be  added  also  to  the  right-hand  side.  Thus 


or 


This  equation  is  precisely  of  the  form  (2),  §  1,  where 
«  =  —  1,  /8  =  3,  p  =  2.  It  therefore  represents  a  circle  whose 
center  is  at  (—  1,  3),  and  whose  radius  is  2. 


THE   CIRCLE  67 

Example  2.     What  curve  is  represented  by  the  equation 

(3)  z2  +  y2  +  l  =  0? 

It  is  clear  that  no  point  exists  whose  coordinates  satisfy  this 
equation.  For,  x*  and  y2  can  never  be  negative.  Their  least 
values  are  0,  —  namely,  for  the  origin,  (0,  0), — and  even  for 
this  point,  the  left-hand  side  of  the  equation  has  the  value  +  1. 
Hence,  there  is  no  curve  corresponding  to  equation  (3). 

Example  3.     Discuss  the  equation 

(4)  a;2  +  y'  +  2a;-4y  +  5  =  0. 
Evidently,  this  equation  can  be  written  in  the  form : 

(5)  (x  +  l)*+(y-  2)^  =  0. 

The  coordinates  of  the  point  (—  1,  2)  satisfy  the  equation. 
But,  for  any  other  point  (x,  y),  at  least  one  of  the  quantities, 
x  -f- 1  and  y  —  2,  is  not  zero,  and  the  left-hand  side  of  the  equa- 
tion is  positive.  Thus  the  point  (—  1,  2)  is  the  only  point 
whose  coordinates  satisfy  the  equation.  Hence  equation  (4) 
represents  a  single  point  ( —  1,  2). 

Remark.  Equation  (5)  can  be  regarded  as  the  limiting  case 
of  the  equation 


when  p  approaches  the  limit  0.  This  equation  represents  a 
circle  of  radius  p  for  all  positive  values  of  p.  When  p  ap- 
proaches 0,  the  circle  shrinks  down  toward  the  point  (—  1,  2) 
as  its  limit.  Accordingly,  equation  (5)  is  sometimes  spoken 
of  as  representing  a  circle  of  zero  radius  or  a  null  circle. 

The  General  Case.  It  is  now  clear  how  to  proceed  in  the 
general  case,  in  order  to  determine  what  curve  equation  (1) 
represents.  The  equation  can  be  written  in  the  form : 

(z2  +  Ax  +  \Ay)  +  (yi  +  By  +  {B2)  =  -  C+^AZ- 
or 

.'KJ+K)^^^ 


68  ANALYTIC   GEOMETRY 

If  the  right-hand  side  is  positive,  i.e.  if 


then  equation  (1)  represents  a  circle,  whose  center  is  at  the 
point  (—  ^A,  —  ^B)  and  whose  radius  is 


If,  however,  A2  +  B2  —  4  C  =  0,  then  equation  (1)  represents 
just  one  point,  namely,  (—  ±A,  —  \B),  —  or,  if  one  prefers,  a 
circle  of  zero  radius  or  a  null  circle. 

Finally,  when  A2  +  B2  —  4  C  <  0,  there  are  no  points  whose 
coordinates  satisfy  (1).  To  sura  up,  then  : 

Equation  (1)  represents  a  circle,  a  single  point,  or  there  is  no 
point  whose  coordinates  satisfy  (1),  according  as  the  expression 

,42  +  52  _  4(7 

is  positive,  zero,  or  negative. 

Consider,  more  generally,  the  equation 
(6)  a(xl  +  y^)  +  bx  +  cy  +  d  —  0. 

If  a  =  0,  but  b  and  c  are  not  both  0,  the  equation  represents 
a  straight  line. 

If,  however,  a  =£  0,  the  equation  can  be  divided  through  by 
a,  and  it  thus  takes  on  the  form  : 


This  is  precisely  the  form  of  equation  (1),  and  hence  the  above 
discussion  is  applicable  to  it. 

EXERCISES  ^ 

Determine  what  the  following  equations  repltesent.  Apply 
each  time  the  method  of  completing  the  square  ai^d  examining 
the  right-hand  side  of  the  new  equation.  Do  not  Merely  sub- 
stitute numerical  values  in  the  formulas  developed  in  the  text. 


THE   CIRCLE 


69 


Ans.   A  circle,  radius  5,  with  center  at  (—3,  4). 

2.  x2  +  f-  —  6x  +  4y  +  13  =  0.       Ans.   The  point  (3,  -  2). 

3.  x1  -f-?/2  -f  2x  +  4:y  +  6  =  0.        -4ns.   No  point  whatever. 

4.  •  a?2  +  y2  —  lOz  +  24  y  =  0. 
y  x"-  +  yz  —  7x=5. 

6.  XB"  +  y"-  —  6ic  +  Sy  +  25  =  0. 

y  494*  +  49y2  -  14  a?  +  28y  +  5  =  0. 

^4ns.    The  point  (^,  —  ^|). 
8. 


12. 

13.  x>  +  ?/-  +  3  =  0. 

14.  x* ,+  y2  —  2x  +  4y  +  10  =  0. 

15.  3cc2  -\-  3yz  —  4#  +  2y  +  7  = 

16.  y5x-  +  5u2  —  60;  +  8w  =  12. 

/ 

IV  3 


gents.     Let  the  circle 


f  (1) 


be  given,  and  let  P!  :  (a?!,  y^)  be  any  point  of  this  circle.     To 
find   the   equation   of    the   tangent 
at  P!. 

The  tangent  at  Px  is,  by  Ele- 
mentary Geometry,  perpendicular  to 
the  radius,  OPi-  Hence  its  slope, 
A',  is  the  negative  reciprocal  of  the 
slope,  t/i/o?!,  of  OP!  ;  or 


FIG.  2 


70  ANALYTIC   GEOMETRY 

We  wish,  therefore,  to  find  the  equation  of  the  line  which 
passes  through  the  point  (xi,  yt)  and  has  the  slope  A/  =  —  x\/y\. 
By  Ch.  II,  §  2,  (II),  the  equation  of  this  line  is 

(2)  y-y^-^x-xj. 

Vi 

This  equation  can  be  simplified  by  multiplying  through  by 
yi  and  transposing  :  ,^  . 

(3)  Bp+Jbf~W  +  )tfr 

Now,  the  point  (xl}  y^  is,  by  hypothesis,  on  the  circle  ;  hence 
its  coordinates  satisfy  the  equation  (1)  of  the  circle  : 

*i2  +  2/i2  =  P2. 

The  right-hand  side  of  equation  (3)  can,  therefore,  be  replaced 
by  the  simpler  expression,  p-. 

We  thus  obtain,  as  the  final  form  of  the  equation  of  the 
tangent,  the  following  : 


In  deducing  this  equation  it  was  tacitly  assumed  that  yl  3=  0, 
since  otherwise  we  could  not  have  divided  by  it  in  obtaining 
X'.  The  final  formula,  (4),  is  true,  however,  even  when  yt  =  0, 
as  can  be  directly  verified.  For,  if  yl  =  0  and  xl  =  p,  then  (4) 
becomes 

px  =  p2        or        x  =  p, 

and  this  is  the  equation  of  the  tangent  in  the  point  (p,  0). 
Similarly,  when  yl  =  0  and  xl  =  —  p. 

Any  Circle.  If  the  given  circle  is  represented  by  the 
equation 

(5)  (s-«)»+(y-/8)l  =  p', 

precisely  the  same  reasoning  can  be  applied.  The  equation 
of  the  tangent  to  (5)  at  the  point  Pl  :  (xly  y^  of  that  circle  is 
thus  found  to  be  :  . 

(6)  (»i-«)(a!-a)  +  (y1-/8)(y-/8)=p«. 
The  proof  is  left  to  the  student  as  an  exercise. 


THE   CIRCLE  71 

If  the  equation  of  the  circle  is  given  in  the  form 

(7)  a2  +  tf  +  Ax  +  By  +  (7=0, 

or  in  the  form  (6),  §  2,  the  equation  can  first  be  thrown  into 
the  form  (5),  an'    then  the  equation  of  the  tangent  is  given 

by  (6). 

Example.     To  find  the  equation  of  the  tangent  to  the  circle 

(8)  3x2  +  3?/2  +  8a;-5?/  =  0 
at  the  origin. 

"First,  reduce  the  coefficients  of  the  terms  in  x-  and  yz  to 
unity  : 


Next,  complete  the  squares  : 


Now,  apply  the  theorem  embodied  in  formula  (6).     Since 

?i  =  0,     2/i  =  0,     «  =  -|,     £  =  f, 
we  have  f(z  +  *)-  fly  -  |)=  If, 

or  8*  -5y  =  0, 

as  the  equation  of  the  tangent  to  (8)  at  the  origin. 


EXERCISES 

Find  the  equation  of  the  tangent  to  each  of  the  following 
circles  at  the  given  point. 

x*  +  y*  =  25    at  (-3,  4).  Ans.   3x  -  ±y  +  25  =  0. 

x2  +  f  =  a2     at  (0,  a).  Ans.   y  =  a. 

x2  +  f-  =  49    at  (-  7,  0). 
(3.  ^  i)2+  (y  +  2)2  =  25     at  (4,  2).       Ans.   3x  +  ±y  =  20. 

5.  (a,  +  5)2+(2/_3)2  =  49    at  (2,  3). 

6.  x2  +  ?/2  — 9a;  +  ll?/  =  0    at  the  origin. 


72  ANALYTIC   GEOMETRY 


V  7 
^3 


7.   2x*  +  2y2-3x-y  =  ll     at  (—1,2).  , 


3.   Find  the  intercepts  on  the  axis  of  x  made  by  the  tangent 
at  (-  5,  12)  to 


169.  /        Ans.    -  33f  . 

9.    Find  the  area  of  the  triangle  cut  from  ,he  first  quadrant 
by  the  tangent  at  (1,  1)  to 

3z2  +  3y2  +  Sx  +  16y  =  30. 
VjO.   If  the  equation 

a2  +  y2  +  Ax  +  By  +  C=  0 

represents  a  circle,  and  if  the  point  (xl9  y^)  lies  on  the  circle, 
show  that  the  equation  of  the  tangent  at  this  point  can  be 
written  in  the  form  : 

(9)  XlX  +  y}y+A(X  +  Xl)+^(y  +  y^+C=Q. 

z  A 

Suggestion.     Find  the  values  of  a,  ft,  and  p  for  the  circle, 
substitute  them  in  (6),  and  simplify  the  result. 

11.  Do  Exs.  6  and  7,  using  formula  (9),  Ex.  10. 

12.  The  same  for  the  tangent  to  the  circle  in  Ex.  9. 

13.  Show  that,  if  Px  :  (xly  yt)  is  any  point  of  the  circle 


at  which  the  tangent  is  not  parallel  to  the  axis  of  y,  then  the 
slope  of  the  tangent  at  Pl  is 


4.  Circle  through  Three  Points.  It  is  shown  in  Elementary 
Geometry  that  a  circle  can  be  passed  through  any  three  points 
not  lying  in  a  straight  line. 

If  the  points  are  (x1}  y^,  (x%,  y2),  and  (x3,  y3),  and  if  the 
equation  of  the  circle  through  them  is  written  in  the  form 


then  clearly  the  following  three  equations  must  hold  : 


THE   CIRCLE  73 

+  Axi  +  %i  +  £  =  0, 


y-22 


We  thus  have  three  simultaneous  linear  equations  for  de- 
termining the  three  unknown  coefficients  A,  B,  C. 

Suppose,  for  example,  that  the  given  points  are  the 
following  : 

(1,  1),         (1,  -  1),         (-  2,  1). 

*  f 

The  equations  can  be  thrown  at  once  into  the  form 

A  +  B  +  C  =  -  2, 

A-B+C  =  -2, 

-2A  +  B+C  =  -5. 

Solve  two  of  these  equations  for  two  of  the  unknowns  in 
terms  of  the  third.  Then  substitute  the  values  thus  found  in 
the  third  equation.  Thus  the  third  unknown  is  completely 
determined,  and  hence  the  other  two  unknowns  can  be  found. 

Here,  it  is  easy  to  solve  the  first  two  equations  for  A  and  B 
in  terms  of  C.  On  subtracting  the  second  equation  from  the 
first,  we  find  : 

2JS  =  0;         hence         B  =  0. 

Then  either  of  the  first  two  equations  gives  for  A  the  value  : 
A  =  -C-2. 

Next,  set  for  A  and  B  in  the  third  equation  the  values  just 
found  : 


Hence,  finally, 

A  =  l,        B  =  Q,        C  =  -3, 

and  the  equation  of  the  desired  circle  is  : 
&  +  y~  +  x  —  3  =  0. 

Check  the  result  by  substituting  the  coordinates  of  the 
given  points  successively  in  this  last  equation.  They  are 
found  each  time  to  satisfy  the  equation. 


74  ANALYTIC  GEOMETRY 

The  circle  through  the  three  given  points  has  its  center  in 
the  point  (-  |,  0).     Its  radius  is  of  length  V&25  =  1.803. 

EXERCISES 

Find  the  equations  of  the   circles   through   the   following 
triples  of  points.     Plot  the  points  and  draw  the  circles. 
•V    (1,  0),  (0,  1),  the  origin.  Ans.   x*  +yz  —  x-  y  =  0. 

2.  (1,1),  (-!,-!),  (1,-1). 

3.  (5,10),  (6,9),  (-2,3). 

4.  The  vertices  of  the  triangle  of  Ex.  15   at   the   end   of 
Ch.  Ill,  p.  63.     Show  that  the  point  (2,  —  1)  of  that  exercise 
lies  on  the  circle. 

5.  The   same   question   for  Ex.  17,  p.  63.     Show  that  <ihe 

point  [0,  —  )  of  that  exercise  lies  on  the  circle. 
V      c  J 

6.  The  vertices  of  the  triangle  of  Ch.  Ill,  Fig.  1.     Find  the 
coordinates  of  the  center  and  check  by  comparing  them  with 
those  of  the  point  of  intersection  of  the  perpendicular  bisectors 
of  the  sides  of  the  triangle,  as  determined  in  Ch.  Ill,  §  5. 

7.  The  same  question  for  the  triangle  of  Ch.  Ill,  Fig.  2. 
Check. 

^J.   The  vertices  of  the  triangle  formed   by  the   coordinate 
axes  and  the  line  2x  —  3y  =  6. 
^9.    The  vertices  of  the  triangle  whose  sides  are  : 


Ans.   3^  +  3^  +  17or 

EXERCISES  ON  CHAPTER   IV 

Nl.    Find  the  equation  of  the  circle  with  the  line-segment  join- 
ing the  two  points  (3,  0)  and  (5,  2)  as  a  diameter. 

2.  A  circle  goes  through  the  origin  and  has  intercepts  —  5 
and  3  on  the  axes  of  x  and  y  respectively.     Find  its  equation. 

3.  A  circle  goes  through  the  origin  and  has  intercepts  a  and 
6.     Find  its  equation. 


THE   CIRCLE  75 

*4.  Find  the  equation  of  the  circle  which  has  its  center  in 
the  point  (—  3,  4)  and  is  tangent  to  the  line  3x  +  8y  —  6  =  0. 

5.    A   circle  has   its   center   on  the  line  2x  —  3y  =  Q  and 
passes  through  the  points  (4,  3),  (—2,  5).     Find  its  equation. 

VB.  Find  the  equation  of  the  circle  which  passes  through  the 
point  (5,  —  2)  and  is  tangent  to  the  line  3x  —  y  —  1  =  0  at 
the  point  (1,  2). 

7.  There  are  two  circles  passing  through  the  points  (3,  2), 
(—  1,  0)  and  having  6  as  their  radius.     Find  their  equations. 

8.  There  are  two  circles  with  their   centers  on  the  line, 
5x  —  3.y  =  8,  and  tangent  to  the  coordinate  axes.     Find  their 
equations. 

Q^.  Find  the  equations  of  the  circles  tangent  to  the  axes  and 
passing  through  the  point  (1,  2). 

10.  Find  the  equations  of  the  circles  passing  through  the 
points  (3,  1),  (1,  0)  and  tangent  to  the  line  x  —  y  =  0. 

Suggestion.  Demand  that  the  center  (a,  j3)  be  equally 
distant  from  the  two  points  and  the  line. 

\I.  Find  the  equations  of  the  circles  passing  through  the 
origin,  tangent  to  the  line  x  -(-  y  —  8  =  0,  and  having  their 
centers  on  the  line  x  =  2. 

12.  Find  the  equations  of  the  circles  of  the  preceding  exer- 
cise, if  their  centers  lie  on  the  line  2x  —  y  —  2  =  0. 

\\13.   Find  the  equation  of  the  circle  inscribed,  in  the  triangle 
formed  by  the  axes  and  the  line  3x  —  4y  —  12  =  0. 

14.  Find  the  equation  of  an  arbitrary  circle,  referred  to  two 
perpendicular  tangents  as  axes. 

15.  Do  the  four  points  (0,  0),  (6,  0),  (0,  -  4),  (5,  1)  lie  on 
a  circle  ? 

16.  Find  the  coordinates  of  the  points  of  intersection  of  the 
circles 


76  ANALYTIC   GEOMETRY 

17.  Find  the  coordinates  of  the  points  of  intersection  of  the 

circles 

2-2  _j_  yi  _|_  ax  _|_  ty  _  o? 

«2  +  y2  +  bx  —  ay  =  0. 

ORTHOGONALITY 

18.  A  circle  and  a  line  intersect  in  a  point  P.     The  acute 
angle  between  the  line  and  the  tangent  to  the  circle  at  P  is 
known  as  the  angle  of  intersection  of  the  line  and  the  circle  at 
P.     If  the  line  meets  the  circle  in  two  points,  the  angles  of 
intersection  at  the  two  points  are  equal.     Determine  the  angle 
in  the  case  of  the  circle 

0^+2/2=25, 

and  the  line  2x  —  y  —  5  =  0. 

19.  A  circle  and  a  line  are  said  to  intersect  orthogonally  if 
their  angle  of  intersection  is  a  right  angle.     Prove  that  the 
circle, 


is  intersected  by  the  line,  5x  +  y  =  7,  orthogonally. 

Suggestion.     First  answer  geometrically  the  question  :  What 
lines  cut  a  given  circle  orthogonally  ? 
-^0.    Show  that  the  circle, 


intersects  the  line, 

ax  +  by  -f  c  =  0, 

orthogonally  when  and  only  when 


21.  If  two  circles  intersect  in  a  point  P,  the  acute  angle 
between  their  tangents  at  P  is  known  as  their  angle  of  inter- 
section. If  the  circles  intersect  in  two  points,  their  angles  of 
intersection  at  these  points  are  equal.  Find  this  angle  in'  the 
case  of  the  circles, 


THE   CIRCLE  77 

22.  Prove  geometrically  that  two  circles  intersect  orthog- 
onally, that  is,  at  right  angles,  when  and  only  when  the  sum 
of  the  squares  of  their  radii  equals  the  square  of  the  distance 
between  their  centers.  Then  show  that  the  circles 


—   2=0, 
2  a2  +  2y°-  +  4  a;  —  6y  -  19  =  0, 

intersect  orthogonally. 
<-^!$.   Prove  that  the  two  circles, 

X2   +   ?/2   +   AiX  +   ^y    +    Ci   =    0, 

a2  +  tf-  +  A2x  +  B2y  +  <72  =  0, 
intersect  orthogonally  when  and  only  when 

A1A2  +  B1B2  =  2  C1+2<72. 
24.    Find  the  equation  of  the  circle  which  cuts  the  circle 


at  right  angles  and  passes  through  the  points  (1,  0)  and  (0,  1). 

25.   There  are  an  infinite  number  of  circles  cutting  each  of 
the  two  circles, 

a?  +  2/2  _  4y  +  2  =  0, 

2  =  0, 


orthogonally.     Show  that  they  are  all  given  by  the  equation 
x2  +  y1  +  ax  -  2  =  0, 

where  a  is  an  arbitrary  constant.     Where  are  their  centers? 
Draw  a  figure. 

26.   Find  the  equation  of  the  circle  cutting  orthogonally  the 
three  circles, 

-  =  9, 


y-19  =  0. 
Ans.   x1-  +  f  +  10z  +  9  =  0. 


78  ANALYTIC  GEOMETRY 

MISCELLANEOUS  THEOREMS 

27.  Prove  analytically  that  every  angle  inscribed  in  a  semi- 
circle is  a  right  angle. 

28.  Prove  analytically  that  the  perpendicular  dropped  from 
a  point  of  a  circle  on  a  diameter  is  a  mean  proportional  between 

,  the  segments  in  which  it  divides  the  diameter. 

29.  The  tangents  to  a  circle  at  two  points  P,  Q  meet  in  the 
point  T.     The  lines  joining  P  and  Q  to  one  extremity  of  the 
diameter  parallel  to  PQ  meet  the  perpendicular  diameter  in 
the  points  R  and  S.     Prove  that  RT=ST. 

30.  In  a  triangle  the  circle  through  the  mid-points  of  the 
sides  passes  through  th'e  feet  of  the  altitudes  and  also  through 
the  points  halfway  between  the  vertices  and  the  point  of  inter- 
section of  the  altitudes.     This  circle  is  known  as  the  Nine- 
Point  Circle  of  the  triangle. 

For  the  triangle  with  vertices  in  the  points  (—4,  0),  (2,  0), 
(0,  6)  construct  the  circle  and  mark  the  nine  points  through 
which  it  passes. 

31.  For  the  triangle  in  the  preceding  exercise  find  the  equa- 
tion of  the  nine-point  circle,  as  the  circle  through  the  mid- 
points of  the  sides.  Ans.   So;2  +  3y7  +  3x  —  lit/  =  0. 

32.  Show  that  this  circle  goes  through  the  other  six  points. 

33.  For  the  triangle  with  vertices  in  the  points  (a,  0),  (6,  0), 
(0,  c)  find  the  equation  of  the  nine-point  circle,  as  the  circle 
through  the  mid-points  of  the  sides. 

Ans.   2c(x>-  +  y°)-(a  +  V)cx  +(ab  -  &)y  =  0. 

34.  Show  that  this  circle  goes  through  the  other  six  points. 


CHAPTER   V 

INTRODUCTORY    PROBLEMS    IN    LOCI.     SYMMETRY    OF 
CURVES 

1.  Locus  Problems.*  A  point  is  moving  under  given  condi- 
tions; its  locus  is  required.  This  type  of  problem  the  student 
studied  in  Plane  Geometry.  But  he  found  there  no  general 
method,  by  means  of  which  he  could  always  determine  a  locus ; 
for  each  problem  he  had  to  devise  a  method,  depending  on  the 
particular  conditions  of  the  problem. 

Analytic  Geometry,  however,  provides  a  general  method  for 
the  determination  of  loci.  Some  simple  examples  of  the 
method  have  already  been  given.  Thus,  in  finding  the  equa- 
tion of  a  circle,  we  determined  the  locus  of  a  point  whose  dis- 
tance from  a  fixed  point  is  constant.  Again,  in  deducing  the 
equation  of  a  line  through  two  points,  we  found  the  locus  of  a 
point  moving  so  that  the  line  joining  it  to  a  given  point  has  a 
given  direction. 

The  method  in  each  of  these  cases  consisted  merely  in  ex- 
pressing in  analytic  terms  —  i.e.  in  the  form  of  an  equation 
involving  the  variable  coordinates,  x  and  y,  of  the  moving 
point  —  the  given  geometric  condition  under  which  the  point 
moved.  We  proceed  to  show  how  this  method  applies  in  less 
simple  cases. 

Example  1.  The  base  of  a  triangle  is  fixed,  and  the  dis- 
tance from  one  end  of  the  base  to  the  mid-point  of  the  opposite 
side  is  given.  Find  the  locus  of  the  vertex. 

*  The  locus  problems  In  this  chapter  may  be  supplemented,  if  it  is  de- 
sired, by  §§  6-8  of  the  second  chapter  on  loci,  Ch.  XIII,  in  which  the  loci 
of  inequalities  and  the  bisectors  of  the  angles  between  two  lines,  together 
with  related  subjects,  are  considered. 

79 


80  ANALYTIC   GEOMETRY 

Let  the  triangle  be  OAP,  with  M  as  the  mid-point  of  AP. 
Let  a  be  the  length  of  the  base  OA,  and  let  I  be  the  given 

distance.     It  is  required  to 
find  the  locus  of  P,  so  that 
always 
(1)  OM  =  I. 

It   is   convenient   to  take 
the  origin  of  coordinates  in 


C  O  A:(a,o)  _.  „ 

0  and  the  positive  axis  of  x 
along  the  base.     The  coordi- 
nates of  A  are  then  (a,  0).     The  coordinates  of  the  moving 
point  P  we  denote  by  (x,  y).     The  coordinates  of  the  point 
Jtfare 


The  distance  OM  is 


a     y\ 

'    2/ 


Thus  condition  (1),  expressed  analytically,  is 


Squaring  both  sides  of  this  equation  and  simplifying,  we  have 
(2)  (x  + 


This  equation  represents  the  circle  whose  center  is  at  (—  a,  0) 
and  whose  radius  is  2  1.  We  have  shown,  therefore,  that,  if 
(1)  is  always  satisfied,  the  coordinates  (x,  y}  of  P  satisfy  (2), 
arid  P  lies  on  the  circle.  The  locus  of  P  appears,  then,  to  be 
the  circle. 

How  do  we  know,  though,  that  P  traces  the  entire  circle  ? 
To  prove  this,  we  must  show,  conversely,  that,  if  the  coordi- 
nates (x,  y)  of  P  satisfy  (2),  condition  (1)  is  valid.  If  (x,  y) 
satisfy  (2),  then,  on  dividing  both  sides  of  (2)  by  4  and  extract- 
ing the  square  root  of  each  side,  we  obtain  two  equations  : 


INTRODUCTORY  PROBLEMS  IN  LOCI 


81 


Equation  ii)  says  that  a  positive  or  zero  quantity  equals  a 
negative  quantity,  and  is  therefore  impossible.  Thus  only 
equation  i)  remains.  This  equation  says  that  OM=  I.  Hence 
condition  (1)  is  satisfied  by  every  point  of  the  circle,*  and  so 
the  circle  is  the  locus  of  P. 

We  have  yet  to  describe  the  locus,  independently  of  the 
coordinate  system,  with  reference  merely  to  the  original  tri- 
angle. Produce  the  base,  in  the  direction  from  A  to  0,  to  the 
point  (7,  doubling  its  length.  Then  the  locus  of  P  is  a  circle, 
whose  center  is  at  C  and  whose  radius  is  twice  the  given 
distance. 

Example  2.  Determine  the  locus  of  a  point  P  which  moves 
so  that  the  difference  of  the  squares  of  its  distances  from  two 
fixed  points  P1}  P2  is  constant, 
and  equal  to  c : 


(3) 


PP22-  PPj2  =  c. 


P:(x,y) 


P,:(a,o) 


FIG.  2 


Take  the  mid-point  of  the 
segment  P\P^  as  origin  and 
the  axis  of  x  along  PiP2. 
The  coordinates  of  P:  and  P2 
can  be  written  as  (—a,  0),  (a,  0)  ;  those  of  P,  as  (x,  y). 

By  Ch.  I,  §  3, 

P?V  =  (a  +  a)2  +  y*,         PP22=  (x  -  a)2  +  y\ 
Then  the  equations  (3),  expressed  analytically,  are 


*  The  two  points  in  which  the  circle  cuts  the  axis  of  x  are  exceptions, 
since  these  do  not  lead  to  a  triangle,  OAP. 


82  ANALYTIC   GEOMETRY 


a)2+  y*—(x  —  a)2  —  y2  =  c, 
(x  -  a)2+  y*  -(x  +  a)2  -y^=c. 
These  reduce  to 

(4)  4  ax  =  c,  4  ax  =  —  c. 

Hence,  if  condition  (3)  is  satisfied,  P  lies  on  one  or  the 
other  of  the  lines 

(5)  *=   '  •—*•?£ 

4a  4a 

Conversely,  if  P  lies  on  one  or  the  other  of  the  lines  (5), 
then  (4)  holds,  and  from  (4)  we  show  by  retracing  the  steps 
that  one  or  the  other  of  the  equations  (3)  is  valid. 

Consequently,  the  locus  of  P  consists  of  two  straight  lines, 
perpendicular  to  the  line  PiP2,  an(i  symmetrically  situated 
with  reference  to  the  mid-point  of  PiP2,  the  distance  of  either 
line  from  the  mid-point  being  c/4  a.  Thus  the  locus  consists 
of  two  entirely  unconnected  pieces,  one  corresponding  to  each 
of  the  equations  (3).  If  c  =  0,  these  equations  are  the  same, 
and  the  two  lines  forming  the  locus  coincide  in  the  perpendic- 
ular bisector  of  the  segment  PiP2. 

EXERCISES 

In  solving  the  following  problems,  the  first  step  is  to  find 
the  equation  of  a  curve,  —  or  the  equations  of  curves,  —  on 
which  points  of  the  locus  lie.  The  student  must  then  take 
care  (a)  to  show,  conversely,  that  every  point  lying  on  the 
curve  or  curves  obtained  satisfies  the  given  conditions  ;  and 

(6)  to  describe   the   locus,  finally,  without   reference   to   the 
coordinate  system  used. 

1.  A  point  P  moves  so  that  the  sum  of  the  squares  of  its 
distances  to  two  fixed  points  Pl}  P2  is  a  constant,  c,  greater 
than  \  PiP22-  Show  that  the  locus  of  P  is  a  circle,  with  its 
center  at  the  mid-point  of  PiP2. 

What  is  the  locus  if  c  =  £  PiP22  ?     If  c  <  | 


INTRODUCTORY   PROBLEMS   IN   LOCI  83 

2.  Find  the  locus  of  the  mid-point  of  a  line  of  fixed  length 
which  moves  so  that  its  end  points  always  lie  on  two  mutually 
perpendicular  lines. 

3.  Determine  the  locus  of  a  point  which  moves  so  that  the 
sum  of  the  squares  of  its  distances  to  the  sides,  or  the  sides 
produced,  of  a  given  square  is  constant.     Is  there  any  restric- 
tion necessary  on  the  value  of  the  constant  ? 

4.  Determine  the  locus  of  a  point  which  moves  so  that  the 
square  of   its   distance   to   the  origin  equals  the  sum  of  its 
coordinates.          Ans.   A  circle,  center  at  (-|,  ^),  radius  =  ^V2. 

5.  Show  that  the  locus  of  a  point  which  moves  so  that  the 
sum   of    its   distances    to   two   mutually   perpendicular   lines 
equals  the  square  of  its  distance  to  their  point  of  intersection 
consists  of  the  arcs  of  four  circles,  forming  a  continuous  curve. 
Where  are  the  circles,  and  which  of  their  arcs  belong  to  the 
locus  ? 

6.  The  base  of  a  triangle  is  fixed,  and  the  trigonometric 
tangent   of  one  base  angle    is  a  constant  multiple,  not  —  1, 
of  the  trigonometric  tangent  of  the  other.     Find  the' locus  of 
the  vertex. 

*  2.    Symmetry.     In  the  problems  of  the  preceding  paragraph, 

the  equations  of  the  loci  were  familiar  and  the  curves  they 

fr  represented  were  easily  identified.     In   subsequent  chapters, 

f\J(r    however,  we  shall  have  locus  problems  to  consider  in  which 

*    the  resulting  equations  will  be  new 
I    I     to  us.     In  drawing  the  curves  which 

these  equations  represent,  it  will  be     (-x-v>- 

useful  to  have  at  hand  the  salient    

facts   concerning   the   symmetry  of 

curves.  ^^ 

Symmetry  in  a  Line.  Two  points, 
P  and  P',  are  said  to  be  symmetric 
in  a  line  L,  if  L  is  the  perpendicular  bisector  of  PP'. 

If  L  is  the  axis  of  x  and  (x,  y)  are  the  coordinates  of  P,  then 
it  is  clear  that  (x,  —  y~)  are  the  coordinates  of  P'. 


84 


ANALYTIC   GEOMETRY 


FIG.  4 


Similarly,  if  L  is  the  axis  of  y  and  P  has  the  coordinates 
(x,  y),  then  P'  has  the  coordinates  (—  x,  y). 

Example  1.   Given  the  curve 
(1)  y*  =  os. 

Let  P :  (a?!,  yi)  be  any  point  on  it,  i.e.  let 

/ON  „  2  />• 

v  y  c/i   ~~    i 

be  a  true  equation.  Then  the  point  P' :  (a^,  —  y^),  symmetric 
to  P  in  the  axis  of  x}  also  lies  on  the  curve. 
For,  if  we  substitute  the  coordinates  of  P'  into 
(1),  the  result  is  (—  y^2  =  x1}  or  (2),  and  (2)  we 
know  is  a  true  equation.  We  say,  then,  that 
the  curve  (1)  is  symmetric  in  the  axis  of  x. 

The  test  for  symmetry  in  the  axis  of  x, 
employed  in  this  example,  is  general  in  appli- 
cation. We  state  it,  and  the  corresponding 
test  for  symmetry  in  the  axis  of  y,  in  the  form 
of  theorems. 

THEOREM  1.     A  curve  is  symmetric  in  the  axis  of  x  if  the  sub- 
stitution of  —y  for  y  in  its  equation  leaves  the  equation  unchanged. 
THEOREM  2.     A  curve  is  symmetric  in  the  axis  of  y  if  the  sub- 
stitution of —x  for  x  in  its  equation  leaves  the  equation  unchanged. 

Symmetry  in  a  Point.  Two  points,  Pand  P',  are  symmetric 
in  a  given  point,  if  the  given  point  is  the  mid-point  of  PP'. 

If    the   given   point  is  the  origin  of 
coordinates  and  P  has  the  coordinates 

(x,  y),  then  the  coordinates  of  P'  are     ; 

evidently  (- x,  —y).  t-*,-v)' 

Example  2.     Consider  the  curve  FlG>  5 

/Q\  „ ™3 

(o)  y  —  x». 

If  P :  (a?!,  2/i)  is  any  point  on  this  curve,  then  the  point 
P' :  (  — &!,— 7/t),  symmetric  to  P  in  the  origin,  is  also  on  the 
curve.  For,  the  condition  that  P'  lies  on  the  curve,  namely, 

—  2/i  =(—  O3         or         —  yl=—  xf, 


INTRODUCTORY   PROBLEMS   IN   LOCI 


85 


FIG.  6 


is  equivalent  to  the  condition  :  y±  =  xf,  that  P  lie  on  the  curve. 
We  say,  then,  that  the  curve  (3)  is  symmetric  in  the  origin. 

This  test,  too,  is  general  in  application ; 
we  formulate  it  as  a  theorem. 

THEOREM  3.  A  curve  is  symmetric  in  the 
origin  of  coordinates,  if  the  substitution  of 
—  x  for  x,  and  of  —  y  for  y,  in  its  equation 
leaves  the  equation  essentially  unchanged. 

A  case  in  which  the  test  leaves  the 
equation  wholly  unchanged  is  that  of  the 
circle,  x2  +  y1  =  p2,  or  the  curve  xy  =  a- 
(Fig-  7). 

Now  the  circle  in  question  is  symmetric  in  both  axes.     It 
follows    then,  without   further   investigation,  that  it  is  sym- 
metric in  the  origin,  the   point  of   intersection  of   the   axes. 
This  conclusion  holds  always ;  in  fact, 
we  may  state  the  theorem. 

THEOREM  4.  If  a  curve  is  symmetric 
in  both  axes  of  coordinates,  it  is  symmetric 
in  the  origin. 

The  details  of  the  proof  are  left  to 
the  student  as  an  exercise.  It  is  to  be 
noted  that  the  converse  of  the  theorem, 
namely,  that  if  a  curve  is  symmetric  in 
the  origin,  it  is  symmetric  in  the  axes,  is 
not  true.  For,  the  curve  of  Example  2 

is  symmetric  in  the  origin,  but  not  symmetric  in  either  axis  ; 
this  is  true  also  of  the  curve  xy  =  a2  of  Fig.  7. 


FIG.  7 


EXERCISES 

1.  Prove  Theorem  4. 

2.  Test,  for  symmetry  in  each  axis  and  in  the  origin,  the 
curves  given  in  the  following  exercises  of  Ch.  I,  §  7  : 

(a)  Exercise  2  ;  (c)  Exercise  7  ; 

(6)  Exercise  6 ;  (d)  Exercise  8. 


86  ANALYTIC  GEOMETRY 

In  each  of  the  following  exercises  test  the  given  curve  for 
symmetry  in  each  axis  and  in  the  origin.     Plot  the  curve. 

3.  xy  +  1i=().  6.    t/2  +  4a;  =  0. 

4.  10y  =  x*.  7.   x*  —  y2=4. 

5.  20x  =  y*.  8.    xz  +  2 y*  =  16. 

EXERCISES  ON   CHAPTER  V 

The  base  of   a   triangle   is   fixed  and  the  ratio   of  the 
^  lengths  of  the  two  sides  is  constant.     Find  the  locus  of  the 
vertex.        Ans.   A  circle,  except  for  one  value  of  the  constant. 
A  point  P  moves  so  that  its  distance  from  a  given  line 
is  proportional  to  the  square  of  its  distance  to  a  given  point 
not  on  Z/.     If  P  remains  always  on  the  same  side  of  L  as 
K,  show  that  its  locus  is  a  circle. 

3.  Find  the  locus  of  P  in  the  preceding  exercise,  if  it  re- 
mains always  on  the  opposite  side  of  L  from  K.     Does  your 
answer  cover  all  cases  ? 

4.  If,  in  Ex.  2,  K  lies  on  L  and  P  may  be  on  either  side  of 
L,  what  is  the  locus  of  P? 

5.  Three  vertices  of  a  quadrilateral   are   fixed.     Find  the 
locus  of  the  fourth,  if  the  area  of  the  quadrilateral  is  constant. 

6.  Find  the  locus  of  a  point   moving  so  that  the  sum  of 
the  squares  of  its  distances  from  the  sides  of  an  equilateral 
triangle  is  constant.     Discuss  all  cases. 

Ans.   A  circle,  center  at  the  point  of  intersection  of  the  me- 
diansj  this  point ;  or  no  locus. 

.X*  •  The  feet  of  the  perpendiculars  from  the  point  P :  (X,  Y) 
on  the  sides  of  the  triangle  with  vertices  in  the  points  (0,  0), 
(3,  0),  (0,  1)  lie  on  a  line.  Find  the  locus  of  P. 

Ans.    The  circle  circumscribing  the  triangle. 

8.  The  preceding  problem,  if  the  triangle  has  the  points 
(2,  0),  (-  3,  0),  (0,  4)  as  vertices. 

9.  Problem   7,  for  the   general   triangle,  with   vertices   at 
(a,  0),  (b,  0),  (0,  c). 


INTRODUCTORY   PROBLEMS   IN   LOCI  87 

10.  Show  that  the  equation  of  the  circle  described  on  the 
line-segment  joining  the  points  (x1}  y^,  (x^,  y%)  as  a  diameter 
may  be  written  in  the  form 

(x  -  x^(x  -x2)  +  (y-  y^(y  -  y2)  =  0. 


Suggestion.  Find  the  locus  of  a  point  P  moving  so  that  the 
two  given  points  always  subtend  at  P  a  right  angle. 

11.  The  two  points,  P  and  P',  are  symmetric  in  the  line, 
x  —  y  =  0,  bisecting  the  angle  between  the  positive  axes  of  x 
and  y.     Show  that,  if  (x,  y}  are  the  coordinates   of   P,   then 
(y,  x)  are  the  coordinates  of  P'. 

12.  Prove  that  a  curve  is  symmetric  in  the  line  x  —  y  =  0 
if  the  interchange  of  x  and  y  in  its  equation  leaves  the  equation 
unchanged. 

13.  If  P  and  P'  are  symmetric  in  the  line  x  +  y  =  0  and 
P  has  the  coordinates  (x,  y),  show  that  the  coordinates  of  P' 
are  (-y,  -  x). 

14.  Give  a  test  for  the  symmetry  of   a   curve  in  the  line 
x  +  y  =  0. 

15.  Test  each  of  the  following  curves  for  symmetry  in  the 
lines  x  —  y  =  0  and  x  +  y  =  0. 

(a)  xy  =  a*;  (c)  x*-  —  yz  =  a2; 

(6)  xy  =  -a-;  (d)  (x-  yf-  2x  -  2y  =  0. 

16.  Plot  the  curve  of  Ex.  15,  (d). 

In  each  of  the  following  exercises  find  the  equation  of  the 
locus  of  the  point  P.  Plot  the  locus  from  the  equation,  mak- 
ing all  the  use  possible  of  the  theory  of  symmetry. 

17.  The  distance  of  P  from  the  line  x  -+-  2  =  0  equals  its 
distance  from  the  point  (2,  0). 

18.  The  sum  of  the  distances  of  P  from  the  points  (3,  0) 
and  (-  3,  0)  is  10. 

19.  The  difference  of  the  distances  of  P  from  the  points 
(5,  0)  and  (-  5,  0)  is  8. 


CHAPTER   VI 


FIG.  1 


THE   PARABOLA 

1.  Definition.  A  parabola  is  defined  as  the  locus  of  a  point 
P,  whose  distance  from  a  fixed  line  D  is  always  equal  to  its 
distance  from  a  fixed  point  F,  not  on 
the  line.  It  is  understood,  of  course, 
that  P  is  restricted  to  the  plane  deter- 
mined by  D  and  F. 

One  point  of  the  locus  is  the  mid-point 
A,  Fig.  2,  of  the  perpendicular  FE 
dropped  from  F  on  D.  Through  A 
draw  T  parallel  to  D.  Then  no  other 
point  on  T,  or  to  the  left  of  T,  can  belong 
to  the  locus,  for  all  such  points  are 
clearly  nearer  to  D  than  they  are  to  F. 
Further  points  of  the  locus  can  be 
obtained  as  follows.  To  the  right  of  T 
draw  L  parallel  to  D,  cutting  AF,  pro- 
duced if  necessary,  in  S.  With  ES  as 
radius  and  F  as  center  describe  a  circle, 
cutting  L  in  P  and  Q.  Then  P  and  Q 
lie  on  the  locus. 

A  large  number  of  points  having  been 
obtained  in  this  way,  a  smooth  curve 
can  be  passed  through  them.  The  curve 
is  symmetric  in  the  line  AF,  and  evi- 
dently has  T  as  a  tangent. 

The  line  D  is  called  the  directrix,  and  the  point  F,  the  focus, 
of  the  parabola ;  A  is  the  vertex,  and  the  indefinite  line  AF, 
the  axis ;  FP  is  a  focal  radius. 

88 


Fia.  2 


THE   PARABOLA 


89 


The  student  is  familiar  with  the  fact  that  all  circles  are 
similar;  i.e.  have  the  same  shape,  and  differ  only  in  size.  A 
like  relation  holds  for  any  two  parabolas.  Think  of  them  as 
lying  in  different  planes,  and  choose  in  each  plane  as  the  unit 
length  the  distance  be- 
tween the  focus  and  the 
directrix.  Then  the  one 
parabola,  in  its  plane,  is 
the  replica  of  the  other, 
in  its  plane.  Conse- 
quently, the  two  parab- 
olas differ  only  in  the 
scale  to  which  they  are 
drawn,  and  are,  there- 
fore, similar. 

The    details    of    the 

proof  just  outlined  can  be  supplied  at  once  by  showing  that 
the  triangles  FPM  and  MFE  are  similar,  respectively,  to 
F'P'M'  and  M'F'E',  the  angles  ^  in  Fig.  3  being  equal  by 
construction.  Hence 

FP  _  EF 
F'P'     E'F'' 

i.e.  focal  radii,  FP  and  F'P',  which  make  the  same  angle  with 
the  axes  always  bear  to  each  other  the  same  fixed  ratio. 


FIG.  3 


EXERCISES 

1.  Take  a  sheet  of  squared  paper  and  mark  D  along  one  of 
the   vertical  rulings   near  the  edge  of  the  paper.     Choose  F 
at  a  distance'  of  1  cm.  from  D.     Then  the  points  of  the  locus 
on  the  vertical  rulings  —  or  on  as  many  of  them  as  one  desires 
—  can  be  marked  off  rapidly  with  the  compasses.     Make  a 
clean,  neat  figure. 

2.  Place  a  card  under  the  curve  of  Ex.  1  and,  with  a  needle, 
prick  numerous  points  of  the  curve  through  on  the  card,  and 
mark,  also,  the  focus  and  axis  in  this  way.     Cut  the  card  along 


90 


ANALYTIC   GEOMETRY 


the  curve  with  sharp  scissors.  The  piece  whose  edge  is  con- 
vex forms  a  convenient  parabolic  ruler,  or  templet,  to  be  used 
whenever  an  accurate  drawing  is  desired. 

A  small  hole  at  the  focus  and  a  second  hole  farther  along 
the  axis  make  it  possible,  in  using  the  templet,  to  mark  the 
focus  and  draw  the  axis. 

A  second  templet,  to  twice  the  above  scale,  will  also  be 
found  useful. 

3.  The  focus  of  a  parabola  is  distant  5  units  from  the 
directrix.  In  a  second  parabola,  this  distance  is  2  units.  How 
much  larger  is  the  first  parabola  than  the  second,  i.e.,  how  do 
their  scales  compare  with  each  other  ? 

2.  Equation  of  the  Parabola.  The  first  step  is  to  choose 
the  axes  of  coordinates  in  a  convenient  manner.  Evidently, 

one  good  choice  would  be  to  take 
the  axis  of  x  perpendicular  to  D 
and  passing  through  F.  Let  us  do 
this,  choosing  the  positive  sense 
from  A  toward  F. 

For  the  axis  of  y  three  simple 
choices  present  themselves,  namely : 

(a)  through  A ; 

(&)  along  D ; 

(c)  through  F. 

Perhaps  (&)  seems  most  natural ; 
but  (a)  has  the  advantage  that  the 
curve  then  passes  through  the 

origin,  and  this  choice  turns  out  in  practice  to  be  the  most 
useful  one.  We  will  begin  with  it. 

Let  P :  (x,  y)  be  any  point  on  the  curve.  Denote  the  dis- 
tance of  F  from  D  by  ra.  Then 


FIG.  4 


=  ~          and 


THE   PARABOLA 
By  Ch.  I,  §  3, 


On  the  other  hand,  the  distance  of  P  from  D  is 


definition,  these  two  distances  are  equal,  or 


Square  each  side  of  the  equation,  so  as  to  remove  the  radical, 
and  expand  the  binomials : 

(2)  x2_ma.  +  ^+2/2  =  a.2  +  ma.  +  ^L2. 

The  result  can  be  reduced  at  once  to  the  form 

(3)  y*  =  2mx, 

and  this  is  the  equation  of  the  parabola',  referred  to  its  vertex 
as  origin  and  to  its  axis  as  the  axis  of  x. 

The  proof  of  this  last  statement  is  not  yet,  however,  com- 
plete ;  for  it  remains  to  show  conversely  that,  if  (x,  y)  be  any 
point  whose  coordinates  satisfy  (3),  it  is  a  point  of  the  parab- 
ola. From  (3)  we  can  pass  to  (2).  On  extracting  the  square 
root  of  each  side  of  (2),  we  have  two  equations : 


one  of  which  must  be  true,  and  both  of  which  may  conceivably 
be  true.  Now,  x  is  a  positive  quantity  or  zero ;  for,  by 
hypothesis,  the  coordinates  of  the  point  (x,  y)  satisfy  equation 
(3).  Hence  ii)  is  impossible,  for  it  says  that  a  positive  or 
zero  quantity  is  equal  to  a  negative  quantity.  Thus  only  i) 


92  ANALYTIC   GEOMETRY 

remains,  and  this  equation  is  precisely  the  condition  that  the 
distance  of  (x,  y)  from  D  be  equal  to  its  distance  from  F. 
Hence  the  point  (x,  y)  lies  on  the  parabola,  q.  e.  d. 


EXERCISES 

1.  Show  that  the  choice  (6)  leads  to  the  equation 

(4)  y2  =  2mx  —  ra2. 

This  is  the  equation  of  the  parabola  referred  to  its  directrix 
and  axis  as  the  axes  of  y  and  x  respectively,  with  the  positive 
axis  of  a;  in  the  direction  in  which  the  curve  opens. 

2.  Show  that  the  choice  (c)  leads  to  the  equation 

(5)  yz  =  2mx  +  ra2. 

This  is  the  equation  of  the  parabola  when  the  focus  is  the 
origin  and  the  positive  axis  of  x  is  along  the  axis  of  the  curve 
in  the  direction  in  which  the  curve 
/  opens. 

P-(x  y\' 

3.  Taking  the  axes  as  indicated  in 
~x  Fig.  5,  show  that  the  equation  of  the 
—• D  parabola  is 

FlG.  5  /vS  —  9  7/3, ?/ 

4.  Choosing  the  axis  of  y  as  in  the  foregoing  question,  show 
that  the  equation  of  the  parabola  is 

xz  =  2my  —  ra2, 
in  case  the  axis  of  x  is  along  D,  and  is 

xz  =  2my  +  ra2, 
in  case  F  is  taken  as  the  origin. 

5.  If  the  axis  of  x  is  taken  along  the  axis  of  the  parabola, 
but  positively  in  the  direction  from  F  toward  D,  and  if  the 
origin  is  taken  at  the  vertex,  show  that  the  equation  of  the 
curve  is 

y2  =  —  2mx. 


7 

4  THE   PARABOLA  93 

6.  If  the  axis  of  y  is  taken  along  the  axis  of  the  parabola, 
but  positively  in  the  direction  from  F  toward  D,  and  if  the 
origin  is  .taken  at  the  vertex,  show  that  the  equation  of  the 

curve  is 

xz  =  —  2my. 

7.  Determine  the  focus  and  directrix  of  each  of  the  follow- 
ing parabolas  : 

(a)  y1-  =  4a;.  Ans.    (1,0);     x  +  1  =  0. 

(6)  y  =  &.  Ans.    (0,  £) ;     4=y  +  1  =  0. 

(c)  3y?  —  5x  =  0.  (d)   3yz  -\ 

(e)  y  =  -2x*.  (/)  5a?H 

8.  It  appears  from  the  foregoing  that  any  equation  of  the 
form 

y~  =  ±  Ax,        or        x1*  =  ±  Ay, 

where  A  is  any  positive  constant,  represents  a  parabola  with 
its  vertex  at  the  origin.  Formulate  a  general  rule  for  ascer- 
taining the  distance  of  the  focus  of  such  a  parabola  from  the 
vertex. 

9.  Find  the  equations  of  the  following  parabolas : 
(a)   Vertex  at  (0,  0)  and  focus  at  (2,  0). 

(&)  Vertex  at  (0,  0)  and  2x  +  5  =  0  as  directrix. 

(c)  Vertex  at  (0,  0)  and  focus  at  (0,  —  |). 

(d)  Vertex  at  (0,  0)  and  2  y  —  1  =  0  as  directrix. 

(e)  Focus  at  (0,  0)  and  vertex  at  (  —  3,  0). 

(/)  Focus  at  (0,  0)  and  3y  +  4  =  0  as  directrix. 
(0)    Focus  at  (6,  0)  and  axis  of  y  as  directrix. 
(Ji)   Focus  at  (0,  —  7)  and  axis  of  x  as  directrix. 

3.  Tangents.  The  student  will  next  turn  to  "Chapter  IX 
and  study  §§  1,  2.  It  is  there  shown  that  the  slope  of  the 
parabola 

(i)  y2  •• 


94  ANALYTIC   GEOMETRY 

at  any  one  of  its  points   (x1}  yt)   is,  in  general,  given  by  the 
formula 

(2)  A  =  ^; 

y\ 

and  that  the  equation  of  the  tangent  line  at  any  point  (x1}  yt) 
can,  without  exception,  be  written  in  the  form 

(3)  yly  =  m(x  +  x1). 

Latus  Rectum.     The  chord,  PP',  of  a  parabola  which  passes 
through  the  focus  and  is  perpendicular  to  the  axis  is  called  the 
latus  rectum  (plural,  latera  recta). 

Its  half-length  is  found  by  setting  x  =  ra/2 
in  the  equation  of  the  parabola,  and  solving 
P:(-%,m)    for  the  positive  y : 


E 


F 


p>  Thus  the  length,  PP1,  of  the  latus  rectum 

is  2m. 

\  The  tangent  at  either  P  or  P'  makes  an 

angle  of  45°  with  the  axis  of  x.     For,  the 
slope  of  the  tangent  at  P  is,  from  (2)  : 

mm 


Let  E  be  the  point  in  which  the  tangent  at  P  meets  the 
axis  of  x.  Since  FP  =  m,  and  Z.FEP  =  45°,  EF=  m  and  so 
E  lies  on  the  directrix.  Consequently,  the  tangents  at  P  and 
P  cut  the  axis  of  x  at  the  point  of  intersection  of  the  directrix 
with  that  axis. 

This  theorem  can  also  be  proved  by  writing  down  the  equa- 
tion of  the  tangent  at  P, 


and  rinding  the  intercept  of  this  line  on  the  axis  of  x. 


THE   PARABOLA  95 

EXERCISES 

1.  Find  the  equation  of  the  tangent  to  the  parabola  y2  =  3x 
at  the  point  (12,  6).  An*,    x  -  4y  +  12  =  0. 

2.  Find  the  equation  of  the  normal  to  the  same  parabola  •/" 
at  the  given  point.  Ans.   4x  +  y  =  54. 

3.  Find  the  length  of  the  latus  rectum  of  the  parabola  of 
Ex.  1. 

4.  Show  that  the  tangents  to  any  parabola  at  the  extremi- 
ties ofthe  latus  rectum  are  perpendicular  to  each  other. 

Show  that  the  tangent  to  the  parabola  y1  =  4  a;  at  the 
point  (36,  12)  cuts  the  negative  axis  of  a;  at  a  point  whose  dis- 
tance from  the  origin  is  36. 

6.    At  what  point  of  the  parabola  of  Ex.  5  is  the  tangent 
perpendicular  to  the  tangent  mentioned  in  that  exercise  ? 


7.    Show  that  the  two  tangents  mentioned  in  Exs.  5  and 
A      6  intersect  on  the  directrix,  and  that  the  chord  of  contact  of 
»    these  tangents,  i.e.  the  right  line  drawn  through  the  two  points 

.^of  tangency,  passes  through  the  focus. 

*\  8.    Show  that  the  tangent  to  the  parabola  (1)  at  any  point 

P  cuts  the  negative  axis  of  x  at  a  point  M  whose  distance 
from  the  origin  is  the  same  as  the  distance  of  P  from  the  axis 
of  y. 

9.   Prove  that  the  two  parabolas, 

2/2  =  4o;  +  4         and        y"'  =  —  60:  +  9, 

intersect  at  right  angles.  Assume  that  the  slope  of  the  parab- 
ola  of  Ex.  2,  §  2,  at  the  point  (xl}  yt)  is  m/y^. 

10.  If  two  parabolas  have  a  common  focus  and  their  axes 
lie  along  the  same  straight  line,  their  vertices,  however,  being 
on  opposite  sides  of  the  focus,  show  that  the  curves  cut  each 
other  at  right  angles. 

4.  Optical  Property  of  the  Parabola.  If  a  polished  reflector, 
like  the  reflector  of  the  headlight  of  a  locomotive  or  a  search- 


96 


ANALYTIC   GEOMETRY 


light,  be  made  in  the  form  of  a  paraboloid  of  revolution,  i.e. 

the  surface  generated  by  a  parabola  which  is  revolved  about 

its  axis,  and  if  a  source  of  light  be  placed  at  the  focus,  the 

reflected  rays  will  all  be  parallel. 

This  phenomenon  is  due  to  the  fact  that  the  focal  radius  FP 

drawn  to  any  point  P  of  the  parabola  makes  the  same  angle 

with  the  tangent  at  P  as  does  the 
line  through  P  parallel  to  the  axis. 
The  proof  of  this  property  can 
be  given  as  follows.  Let  the 
tangent  at  P :  (x^  y^)  cut  the  axis 
of  a;  in  M.  Then  the  length  of 
OM  is  equal  to  x±,  by  §  3,  Ex.  8. 
Furthermore,  OF  =  m/2.  Hence 
the  distance  from  M  to  F  is 


D 


FIG.  7 

But  this  is  precisely  the  distance  of  P  from  D,  §  2,  and 
hence,  by  the  definition  of  the  parabola,  it  is  also  equal  to  FP. 
We  have,  then,  that  MF=FP.  Consequently,  the  triangle 
MFP  is  isosceles,  and 

£  FMP  =  4  MPF. 
But  %FMP=%SPT, 

and  the  proposition  is  proved. 

The  result  can  be  restated  in  the  following 

THEOKEM.  The  focal  radius  FP  of  a  parabola  at  any  point 
P  of  the  curve  and  the  parallel  to  the  axis  at  P  make  equal  angles 
with  the  tangent  at  P. 

Heat.  If  such  a  parabolic  reflector  as  the  one  described 
above  were  turned  toward  the  sun,  the  latter's  rays,  being 
practically  parallel  to  each  other  and  to  the  axis  of  the  reflector, 
would,  after  impinging  on  the  polished  surface,  proceed  along 
lines,  all  of  which  would  pass  through  F.  .Thus,  in  particu- 
lar, the  heat  rays  would  be  collected  at  F,  and  if  a  minute 


THE   PARABOLA  97 

charge  of  gunpowder  were  placed  at  F,  it  might  easily  be 
tired. 

It  is  to  this  property  that  the  focus  (German,  Brennpunkt) 
owes  its  name.  The  Latin  word  means  hetirth,  or  fireplace. 
The  term  was  introduced  into  the  science  by  the  astronomer 
Kepler  in  1604. 

EXERCISES   ON   CHAPTER  VI 

1.  A  parabola  opens  out  along  the  positive  axis  of  y  as  axis. 
Its  focus  is  in  the  point  (0,  3)  and   the   length  of   its    latus 
rectum  is  12.     Find  its  equation.  Ans.    x2  =  12 y. 

2.  A  parabola  has  its  vertex  in  the  origin  and  its  axis  along 
the  axis  of  x.    If  it  goes  through  the  point  (2,  —  3),  what  is  its 
equation  ?  .4ns.   2?/2  —  9  a;  =  0. 

3.  Show  that  the  equation  of  a  parabola  with  the  line  x  =  c 
as  directrix  and  with  the  point   (c  +  m,  0)    or   (c  —  ra,  0)   as 
focus  is 

yz  =  2m(x  —  c)  —  m2,         or        y1*  =  —  2m(x  —  c)  —  m2. 

Hence  prove  that  every  parabola  with  the  axis  of  x  as  axis 
has  an  equation  of  the  form  :  x  =  ay1  +  6,  where  a  and  b  are 
constants,  a  =£  0. 

4.  Find  the  equation  of   the  parabola  which  has   its   axis 
along  the  axis  of  x  and  goes  through  the  two  points  (3,  2), 
(_2,  -1).  Ans.   3x  =  5y*-ll. 

5.  Prove  that  every  parabola  with  an  axis  parallel  to  the 
axis  of  y  has  an  equation  of  the  form 

y  =  ax2  +  bx  +  c, 

where  a,  6,  c  are  constants,  a  =£  0. 

Suggestion.  Find  the  equation  of  the  parabola  which  has 
the  line  y  =  k  as  directrix  and  the  point  (I,  k  +  m)  or  (I,  Jc  —  m) 
as  focus. 

6.  Find  the  equation  of  the  parabola  which  has  a  vertical 
axis  and  goes  through  the  points  (0,  0),  (1,  0),  and  (3,  6). 

Ans.   y  =  xl  —  x. 


98  ANALYTIC   GEOMETRY 

7.  A  circle  is  tangent  to  the  parabola  y1  —  x  at  the  point 
(4,  2)  and  goes  through  the  vertex  of  the  parabola.     Find  its 
equation. 

8.  What  is  the  equation  of  the  circle  which  is  tangent  to 
the  parabola  y-  =  2mx  at  both  extremities  of  the  latus  rectum  ? 

Ans.   4<c2  +  4y2  —  12mo;  +  ra2  =  0. 

9.  Find  the  coordinates  of  the  points  of  tangency  of  the 
tangents  to  the  parabola  y -  =  2  mx  which  make  the  angles  60°, 
45°,  and  30°  with  the  axis  of  the  parabola.     Show  that  the 
abscissae  of  the  three  points  are  in  geometric  progression,  and 

\    that  this  is  true  also  of  the  ordinates. 

A  10.  Show  that  the  common  chord  of  a  parabola,  and  the 
circle  whose  center  is  in  the  vertex  of  the  parabola  and  whose 
radius  is  equal  to  three  halves  the  distance  from  the  vertex  to 
the  focus,  bisects  the  line-segment  joining  the  vertex  with  the 
focus. 

11.  Let  N  be  the  point  in  which  the  normal  to  a  parabola 
at  a  point  P,  not  the  vertex,  meets  the  axis.     Prove  that  the 
projection  on  the  axis  of  the  line-segment  PN  is  equal  to  one 
half  the  length  of  the  latus  rectum. 

12.  On  a  parabola,  P  is  any  point  other  than  the  vertex, 
and  N  is  the  point  in  which  the  normal  at  P  meets  the  axis. 
Show  that  P  and  N  are  equally  distant  from  the  focus. 

13.  The  tangent  to  a  parabola  at  a  point  P,  not  the  vertex, 
meets  the  directrix  in  the  point  L.     Prove  that  the  segment 
LP  subtends  a  right  angle  at  the  focus. 

14.  Show  that  the  length  of  a  focal  chord  of  the  parabola 
y1  =  2  mx  is  equal  to  xt  +  xz  +  ra,  where  xl}  x2  are  the  abscissae 
of  the  end-points  of  the  chord.     Hence  show  that  the  mid- 
point of  a  focal  chord  is  at  the  same  distance  from  the  direc- 
trix as  it  is  from  the  end-points  of  the  chord. 

Exercises  15-26.  The  following  exercises  express  properties 
of  the  parabola  which  involve  an  arbitrary  point  on  the  parab- 
ola. In  order  to  prove  these  properties,  it  will,  in  general,  be 


THE    PARABOLA  99 

necessary  to  make  actual  use  of  the  equation  which  expresses 
analytically  the  fact  that  the  point  lies  on  the  parabola. 

15.  An  arbitrary  point  P  of  a  parabola,  not  the  vertex,  is 
joined  with  the  vertex  A,  and  a  second  line  is  drawn  through 
P,  perpendicular  to  AP,  meeting  the  axis  in  Q.     Prove  that 
the  projection  on  the  axis  of  PQ  is  equal  to  the  length  of  the 
latus  rectum. 

16.  The  tangent  to  a  parabola  at  a  point  P,  not  the  vertex, 
meets  the  tangent  at  the  vertex  in  the  point  K.     Show  that 
the  line  joining  K  to  the  focus  is  perpendicular  to  the  tangent 
at  P. 

17.  The  tangent  to  a  parabola  at  a  point  P,  not  the  vertex, 
meets  the  directrix  and  the  latus  rectum  produced  in  points 
which    are    equally    distant    from    the    focus.      Prove    this 
theorem. 

18.  Prove  that  the  coordinates  of  the  point  of  intersection 
of  the  tangents  to  the  parabola  yz=2  mx  at  the  points  (x1}  y^, 
(fy)  y-t)  mav  be  put  in  the  form 

(Ml      Ih  +  y*\ 
\2m'         2     j 

Suggestion.  To  reduce  the  coordinates  to  the  desired  form, 
use  the  equations  which  express  analytically  the  fact  that  the 
two  points  lie  on  the  parabola. 

19.  Show  that  the  intercept  on  the  axis  of  x  of  the  line  join- 
ing the  points  (»u  yx),  (*j,  y2)  of  the  parabola  y2  =  2  mx  may  be 
expressed  as 


2m 

By  means  of  the  results  of  the  two  preceding  exercises  prove 
the  following  theorems. 

20.  The  point  of  intersection  of  two  tangents  to  a  parabola 
and  the  point  of  intersection  with  the  axis  of  the  line  joining 
their  points  of  contact  are  equally  distant  from  the  tangent  at 
the  vertex,  and  are  either  on  it  or  on  opposite  sides  of  it. 


100  ANALYTIC    GEOMETRY 

21.  Tangents   to  a  parabola  at  the   end-points  of  a  focal 
chord  meet  at  right  angles  on  the  directrix. 

22.  If  the  points  of  contact  of  two  tangents  to  a  parabola 
are  on  the  same  side  of  the  axis  and  at  distances  from  the  axis 
whose  product  is  the  square  of  half  the  length  of  the  latus 
rectum,  the  tangents  intersect  on  the  latus  rectum  produced. 

23o  The  end-points  of  a  chord  of  a  parabola,  which  sub- 
tends a  right  angle  at  the  vertex,  are  on  opposite  sides  of  the 
axis  and  at  distances  from  the  axis,  whose  product  is  the 
square  of  the  length  of  the  latus  rectum. 

24.  The  chords  of  a  parabola,  which  subtend  a  right  angle 
at  the  vertex,  pass  through  a  common  point  on  the  axis ;  this 
point  is  at  a  distance  from  the  vertex  equal  to  the  latus  rectum. 

25.  The  distance  from  the  focus  of  a  parabola  to  the  point 
of  intersection  of  two  tangents  is  a  mean  proportional  between 
the  focal  radii  to  the  points  of  tangency. 

26.  The  tangents  to  a  parabola  at  the  points  P  and  Q  inter- 
sect in  T,  and  the  normals  at  P  and  Q  meet  in  N.     Then  the 
segment  TM,  where  M  is  the  mid-point  of    TN,  subtends  a 
right  angle  at  the  focus. 

Locus  PROBLEMS 

27.  Show  that  the  locus  of  a  point  which  moves  so  that 
the  difference  of  the  slopes  of  the  lines  joining  it  to  two  fixed 
points  is  constant  is  a  parabola  through  the  two  fixed  points. 
What  are  its  axis  and  vertex  ? 

28.  Determine  the  locus  of  a  point  which  moves  so  that  its 
distance  from  a  fixed  circle  equals  its  distance  from  a  fixed 
line  passing  through  the  center  of  the  circle. 

Ans.  Two  equal  parabolas,  with  foci  at  the  center  of  the 
circle  and  axes  perpendicular  to  the  fixed  line. 

29.  The  base  of  a  triangle  is  fixed  and  the  sum  of  the  trigo- 
nometric tangents  of  the  base  angles  is  constant.     Find  the 
locus  of  the  vertex. 


CHAPTER   VII 
THE   ELLIPSE 

1.  Definition.  An  ellipse  is  defined  as  the  locus  of  a  point 
P,  the  sum  of  whose  distances  from  two  given  points,  F  and 
F',  is  constant.  It  is  found  con- 
venient  to  denote  this  constant 
by  2  a.  Then 

(1)  FP+F'P  =  2a. 

It  is  understood,  of  course,  that  P 
always  lies  in  a  fixed  plane  pass- 
ing through  F  and  F'.  ~~~Fm"T 

The  points  F  and  F'  are  called 

the  foci  of  the  ellipse.     It  is  clear  that  2  a  must  be  greater 
than  the  distance  between  them. 

Mechanical  Construction.  From  the  definition  of  the  ellipse 
a  simple  mechanical  construction  readily  presents  itself.  Let 
a  string,  of  length  2  a,  have  its  ends  fastened  at  F  and  F',  and 
let  the  string  be  kept  taut  by  a  pencil  point  at  P.  As  the 
pencil  moves,  its  point  obviously  traces  out  on  the  paper  the 
ellipse. 

The  student  will  find  it  convenient  to  use  two  thumb  tacks 
partially  inserted  at  F  and  F'.  A  silk  thread  can  be  tied  to 
one  of  the  thumb  tacks  and  wound  round  the  other  so  that 
it  will  not  slip.  Thus  a  variety  of  ellipses  with  different  foci 
and  different  values  of  a  can  be  drawn. 

Let  the  student  make  finally  one  ellipse  in  this  manner,  and 
draw  it  neatly. 

101 


102 


ANALYTIC   GEOMETRY 


Center,  Vertices,  Axes.  It  is  obvious  from  the  definition,  — 
and  the  fact  becomes  more  striking  from  the  mechanical .  con- 
struction, —  that  the  ellipse  is  symmetric  in  the  line  through 
the  foci.  It  is  also  symmetric  in  the  perpendicular  bisector 
of  FF'.  Hence  it  is  symmetric,  furthermore,  in  the  mid-point, 

0,  of  the  line  FF'. 

The  indefinite  line  through 
the  foci,  F  and  F',  is  called  the 
transverse  axis  of  the  ellipse ; 
the  perpendicular  bisector  of 
FF',  the  conjugate  axis.  The 
point  0  is  called  the  center  of 
the  ellipse ;  the  points  A,  A', 
its  vertices. 

The  line-segments  AA'  and 
BB',  which  measure  the  length  and  breadth  of  the  ellipse, 
are  known  respectively  as  the  major  axis  and  the  minor 
axis  of  the  ellipse.  The  word  "  axes "  refers  sometimes  to 
the  transverse  and  conjugate  axes,  and  sometimes  to  the  major 
and  minor  axes,  or  their  lengths,  the  context  making  clear  in 
any  case  the  meaning. 

When  P  is  at  A,  equation  (1)  becomes 


But 
Hence 


FA  =  A'F1. 

and         OA  =  a. 


Thus  it  appears  that  the  length  of  the  semi-axis  major,  OA, 
is  a.     Let  the  length  of  the  minor  axis  be  denoted  by  26,  and 
the  distance  between  the  foci  by  2c.     Then,  from  the  triangle 
FOB,  we  have : 
(2)  a?  =  V  +  c2. 

Note  that,  of  the  three  quantities  a,  b,  and  c,  the  quantity  a 
is  always  the  largest. 

Eccentricity.     All  circles  have  the  same  shape,  i.e.  are  simi. 
lar ;  and  the  same  is  true  of  parabolas.     But  it  is  not  true  of 


THE    ELLIPSE  103 

ellipses.  As  a  measure  of  the  roundness  or  flatness  of  an 
ellipse  a  number,  called  the  eccentricity,  has  been  chosen  ;  this 
number  is  defined  as  the  radio  c/a  and  is  denoted  by  e  : 

(3)^  .-£. 

>  Since  c  is  always  less  than  a,  it  is  seen  that  the  eccentricity 
orVan  ellipse  is  always  less  than  unity  : 


In  terms  of  a  and  b,  e  has  the  value  : 

(4)  c^y^y2 

All  ellipses  with  the  same  eccentricity  are  similar,  and  con- 
versely. For  the  shape  of  an  ellipse  depends  only  on  b/a,  the 
ratio  of  its  breadth  to  its  length,  and  since  from  (4) 


all  ellipses  for  which  the  ratio  b/a  is  the  same  have  the  same 
eccentricity,  and  conversely. 

A  circle  is  the  limiting  case  of  an  ellipse  whose  foci  ap- 
proach each  other,  the  length  2  a  remaining  constant.  The 
eccentricity  approaches  0,  and  a  circle  is  often  spoken  of  as  an 
ellipse  of  eccentricity  0. 

EXERCISES 

1.  The  semi-axes  of  an  ellipse  are  of  lengths  3  cm.  and  5  cm. 
Find  the  distance  between  the  foci,  and  the  eccentricity. 

Ans.    8  ;  f . 

2.  The   eccentricity  of  an   ellipse   is   £  and  the  semi-axis 
minor  is  4  in.  long.     How  long  is  the  major  axis  ? 

3.  The  major  axis  of  an  ellipse  is  twice   as   great  as  the 
minor  axis.     What  is  the  eccentricity  of  the  ellipse  ? 

4.  The  major  axis  of  an  ellipse  is  39  yards,  and  the  eccen- 
tricity, y5^.     Find  the  minor  axis. 


104  ANALYTIC   GEOMETRY 

5.  Express  the  eccentricity  of  an  ellipse  in  terms  of  b  and  c. 

6.  Show,  from  Fig.  2,  that  the  eccentricity  is  given  by  the 

formula 

e  =  cos  OFB. 

7.  Give  a  proof,  based  on  similar  triangles,  that  two  ellipses 
having  the  same  eccentricity  are  similar. 

2.   Geometrical  Construction.     Points  on  the  ellipse  may  be 
obtained  with   speed   and   accuracy  by  a  simple   geometrical 

construction.     Draw  the  major 

~^~  axis  and  mark  the  points  A,  F, 

F',  A'  on  it.     Mark  an  arbitrary 

—  *-,  -  H-  -  f  —  i  —  i—    point    Q    between  F  and   F'. 

£  Jp'  W        A 

With  F  as  center  and  AQ  as 

.  ,  radius    describe   a    circle,   and 

FlG  3  with  F'  as  center  and  A'Q  as 

radius  describe  a  second  circle. 

The  points  of  intersection  of  these  two  circles  will  lie  on  the 
ellipse,  since  the  sum  of  the  radii  is 


'Q=2a. 

It  is,  of  course,  not  necessary  to  draw  the  complete  circles, 
but  only  so  much  of  them  as  to  determine  their  points  of  in- 
tersection. Moreover,  four  points,  instead  of  two,  can  be  ob- 
tained from  each  pair  of  settings  of  the  compasses  by  simply 
reversing  the  roles  of  F  and  F'. 

EXERCISES 

1.  Construct  the  ellipse  for  which  c  =  21  cm.,  a  =  4  cm. 

2.  From  the  ellipse  just  constructed  make  a  templet,  with 
holes  at  the  foci  and  with  the  axes  properly  drawn. 

3.  Construct  the  ellipse  whose  axes  are  4  cm.  and  6  cm. 

3.  Equation  of  the  Ellipse.  It  is  natural  to  choose  the  axes 
of  the  ellipse  as  the  coordinate  axes  (Fig.  4).  Let  the  foci  lie 


THE   ELLIPSE 


105 


on  the  axis  of  x,  and  let  P :   (x,  y)  be  any  point  of  the  ellipsa 
Then,  from  (1),  §  1, 

v 


FIG.  4 


(1)  V(»  -  c)2+  2/2  +  V(a?  +  c)2+  2/2  =  2  a. 

Transpose  one  of  the  radicals  and  square : 


(x  —  c)2+  yz=  (x  +  c)2+  y2  —  4  a\(x  +  c)2  +  y*  +  4  a1. 
Hence 

(2)  a  V(a?  +  c)2  +  ^2  =  a2  +  ca. 
To  remove  this  radical,  square  again : 

(3)  a2»2  +  2  a2ca  +  a2c2  +  a2?/2  =  a4  +  2  a2cx  +  c2z2, 
or  (a2  —  c2)o;2  +  a2?/2  =  a2(a2  -  c2). 

But,  by  (2),  §  1,  a2  -  c2  =  62, 

and  hence 

(4)  62a;2  +  a2?/2  =  a26*, 
or 

(5)  ^  +  ^=1. 

a2     62 

This  is  the  standard  form  of  the  equation  of  the  ellipse,  re- 
ferred to  its  axes  as  the  axes  of  coordinates.  The  proof,  how- 
ever, is  not  as  yet  complete,  for  it  remains  to  show,  conversely, 
that  any  point  (x,  y)  whose  coordinates  satisfy  equation  (5) 
is  a  point  of  the  ellipse.  To  do  this,  it  is  sufficient  to  show 
that  x,  y  satisfy  (1).  From  (5)  we  mount  up  to  (4)  and  thence 
to  (3),  since  all  of  these  are  equivalent  equations.  When, 


106  ANALYTIC   GEOMETRY 

however,  we   extract  a  square  root  we   obtain  two  equations 
each  time,  and  so  we  are  led,  finally,  to  the  four  equations 

±  V(«  —  c)2+  y*±  V(ar  +  c)2+  y*  =  2  a, 

the  ambiguous  signs  being  chosen  in  all  possible  ways.     The 
four  equations  can  be  characterized  as  follows  : 

i)         +     +5  ii)         +      -; 

iii)         -     +;  iv)         -      -. 

We  wish  to  show  that  i)  is  the  only  possible  one  of  the  four 
equations.  This  is  done  as  follows. 

Equation  iv)  is  satisfied  by  no  pair  of  values  for  x  and  y, 
since  the  left-hand  side  is  always  negative  and  so  can  never  be 
equal  to  the  positive  quantity  2  a. 

Equations  ii)  and  iii)  say  that  the  difference  of  the  distances 
of  (x,  y)  from  F  and  F'  is  equal  to  2  a,  and  hence  greater  than 
the  line  FF'=  2  c.  Thus,  in  the  triangle  FPF'  the  difference 
of  two  sides  is  greater  than  the  third  side,  and  this  is  absurd.* 
Hence  equations  ii)  and  iii)  are  impossible  and  equation  i) 
alone  remains,  q.  e.  d. 

Consequently,  if  we  start  with  equation  (5)  as  given  and 
require  that  a  >  b,  then  (5)  represents  an  ellipse  with 
semi-axes  a  and  b  and  foci  in  the  points  (±  c,  0),  where 
c=Va2-6?. 

The  Focal  Radii.  From  equation  (2)  we  obtain  a  simple 
expression  for  the  length  of  the  focal  radius,  F'P.  Dividing 
(2)  by  a  and  remembering  that  c/a  =  e,  we  have,: 


V(o;  +  c)2  +  y1  =•  a  +  ex. 

But  the  value  of  the  left-hand  side  of  this  equation  is  precisely 

F'P.     Hence 

(6)  F'P=a  +  ex. 

*  If,  in  particular,  the  point  (x,  y)  lay  on  FF',  we  should  not,  it  is 
true,  have  a  triangle.  But  it  is  at  once  obvious  that  in  this  case,  too, 
equations  ii)  and  iii)  are  impossible. 


THE    ELLIPSE  107 

If,  in  transforming  (1),  the  other  radical  had  been  transposed 
to  the  right-hand  side  and  we  had  then  proceeded  as  before, 
we  should  have  found  the  equation : 


a~\/(x  —  c)2  +  yl  —  a?  —  ex.  , 
From  this  we  infer  that 


V(#  —  c)2  +  y2  =  a  —  ex, 
or 
(7)  FP=a-ex. 

EXERCISES 

1.  What  is  the  equation  of  the  ellipse  whose  axes  are  of 

X1'        V2 

lengths  6  cm.  and  10  cm.  ?  Ans. \-  —  =  1. 

2.  Find  the  coordinates  of  the  foci  of  the  ellipse  of  Ex.  1.       .    _ 
<^)  The  foci  of  an  ellipse  are  at  the  points  (1,  0)  and  (—1,0),  ^ 

and  the  minor  axis  is  of  length  2.     Find  the  equation  of  the 
ellipse.  Ans.   y?  +  2yz  =  2.Q    ~^ 

4.  Find  the  lengths  of  the  axes,  the  coordinates  of  the  foci, 
and  the  eccentricity  of  the  ellipse  ;   .*-  ' 

25  y?  +  169  y2  =  4225. 

5.  An  ellipse,  whose  axes  are  of  lengths  8  and  10,  has  its 
center  at  the  origin  and  its  foci  on  the  axis  of  y.     Obtain  its 
equation. 

6.  Show  that,  if  B  >  A,  the  equation 


still  represents  an  ellipse  with  its  axes  lying  along  the  axes  of 
coordinates  ;  but  the  foci  lie  on  the  axis  of  y  at  the  points 
(0,  C)  and  (0,  -  C),  where 

B2  =  A-  +  C2. 
The  eccentricity  is 


108 


ANALYTIC   GEOMETRY 


7.  Find  the  lengths  of  the  axes,  the  coordinates  of  the  foci, 
and  the  value  of  the  eccentricity  for  each  of  the  following 
ellipses : 

(a)  9x2  +  42/2  =  36;  (d)     5z2  +  3*/2  =  45 ; 

"2  =  19-  (e)     2z2+72/2  =  10; 

(/)  Ilx2  +    y2=    3. 

\t/  J  *          \7 

lr 

'{Tangents.     The  ellipse  has  the  remarkable  property  that 
tangent  to  the  curve  at  any  point  makes  equal  angles  with 
the  focal  radii  drawn  to  that  point : 


JIG.  5 


i)    MecJianical    Proof.      The    simplest 
proof  of  this   theorem   is  a  mechanical 
one.     Think  of  a  flexible,  inelastic  string 
of  length  2  a  with  its  ends  fastened  at 
the  foci,  F  and  F'.      Suppose  a  small, 
smooth  bead  to  be  threaded  on  this  string.     Let  a  cord  be 
fastened  to  the  bead  and  then  pulled  taut,  so  that  the  cord 
and  the  two  portions  of   the  string   will   be  under  tension. 
Evidently,  the  bead  can  be  held  in  this  manner  at  any  point. 
(No  force  of  gravity  is  supposed  to  act.     The  strings  and  bead 
may  be  thought  of  as  resting  on  a  smooth  horizontal  table.) 
The  forces  that  act  on  the  bead  are : 
(a)  the  tension  S  in  the  cord ; 
(6)  two  equal  tensions,  R,  in  the  string, 
directed  respectively  toward  the  foci.* 

Draw  the  parallelogram  of  forces  for  the 
forces  R.  It  will  be  a  rhombus,  and  so 
the  resultant  of  these  forces  will  bisect 
the  angle  between  the  focal  radii. 

On  the  other  hand,  the  force  S,  equal  and  opposite  to  this 
resultant,  is  perpendicular  to  the  tangent  at  P.     In  fact,  if 

*  Since  the  bead  is  smooth,  the  tension  in  the  string  is  the  same  at  all 
its  points,  and  so,  in  particular,  is  the  same  on  the  two  sides  of  the  bead. 


FIG.  6 


&   o 

THE   ELLIPSE  109 

instead  of  the  flexible  string  we  had  a  smooth  rigid  wire,  in  the 
form  of  the  ellipse,  for  the  bead  to  slide  on,  the  bead  would  be 
held  at  P  by  the  cord  exactly  as  before.  But  the  reaction  of 
a  smooth  wire  is  at  right  angles  to  its  tangent.  This  is  the 
very  conception  of  a  smooth  wire.  For  otherwise,  if  S  were 
oblique,  it  could  be  resolved  into  a  normal  and  a  tangential 
component.  But  the  smooth  wire  could  not  yield  a  reaction, 
part  of  which  is  along  the  tangent. 

It  follows,  then,  that  the  normal  at  P  bisects  the  angle  be- 
tween the  focal  radii,  and  hence  these  make  equal  angles  with 
the  tangent  at  P,  q.  e.  d. 

ii)    Proof  by  Means  of  Minimum  Distances.     A  Lemma.     A 
barnyard  is  bounded  on  one  side   by  a  straight   river.     The 
cows,  as  they  come  from  the  pasture, 
enter  the  barnyard  by  a  gate  at  A, 
go  to  the  river  to  drink,  and  then 
keep  on  to  the  door  of  the  barn  at  B. 
What  point,  P,  of  the  river  should  a 


cow  select,  in  order  to  save  her  steps  FlG  7 

so  far  as  possible  ? 

It  is  easy  to  answer  this  question  by  means  of  a  simple  con- 
struction. From  B  drop  a  perpendicular  BM  on  the  line  of 
the  river  bank,  L,  and  produce  it  to  B',  making  MB'  =  BM. 
Join  A  with  B',  and  let  AB'  cut  L  at  C.  Then  C  is  the  posi- 
tion of  P,  for  which  the  distance  under  consideration, 

AP  +  PB, 

is  least. 

For,  the  straight  line  AB'  is  shorter  than  any  broken  line 
APB': 

AB'<  APB'. 

But  PB  =  PB'         and         CB  =  CB'. 

Hence 

A  B'  =  A  C  +  CB        and         APB'  =  AP  +  PB. 

It  follows,  then,  that 

AC+CB<AP+  PB, 


110  ANALYTIC   GEOMETRY 

if  P  is  any  point  of  L  distinct  from  C.     Hence  C  is  the  point 
for  which  APB  is  a  minimum. 

The  point  C  is  evidently  characterized  by  the  fact  that 


We  can  state  the  result,  then,  by  saying  that  the  point  P,  for 
which  the  distance  APB  is  least,  is  the  point  for  which 


Optical  Interpretation.  We  have  used  a  homely  example  of 
cows  and  a  barnyard.  The  problem  we  have  solved  is,  however, 
identical  with  the  optical  problem  of  finding  the  point  at 
which  a  ray  of  light,  emanating  from  A,  will  strike  a  plane 
mirror  L,  if  the  reflected  ray  is  to  pass  through  B.  For,  the 
law  of  light  is,  that  it  will  travel  the  distance  in  the  shortest 
possible  time,  and  hence  it  will  choose  the  shortest  path. 

Application  to  the  Ellipse.     The  application  of  this  result 
to  the  ellipse  is  as  follows.     The  tangent  to  any  smooth,  closed, 
convex  curve  evidently  is  characterized 
by  the  fact  that  it  meets  the  curve  in 
one,  and  only  one,  point. 

Let  P  be  any  point  of  the  ellipse. 
Draw  the  tangent,  T,  at  P.  Let  Q  be 
any  point  of  T  distinct  from  P.  Now 

F'P  +  FP  =  F'R  +  PR, 

since  the  sum  of  the  focal  radii  is  the  same  for  all  points  of  an 
ellipse.     But 

FR<RQ  +  FQ 
and  so 

F'R  +  FM<  F'R  +  RQ  +  FQ  =  F'Q  +  FQ. 

Therefore 

F'P  +  FP<  F'Q  +  FQ. 

Hence  Pis  that  point  of  T  for  which  the  distance  F'QF  is 
least,  and  consequently  the  lines  F'P  and  FP  make  equal 
angles  with  T,  q.  e.  d. 


THE    ELLIPSE  111 

EXERCISE 

Show  that  the  normal  of  an  ellipse  at  any  point  distinct  from 
the  vertices  A,  A'  cuts  the  major  axis  at  a  point  which  lies 
between  the  foci. 

5.  Optical  and  Acoustical  Meaning  of  the  Foci.  Let  a  thin 
strip  of  metal,  —  say,  a  strip  of  brass  a  yard  long  and  a  quarter 
of  an  inch  wide,  —  be  bent  into  the  form  of  an  ellipse  and 
polished  011  the  concave  side.  Let  a  light  be  placed  at  one  of 
the  foci.  Then  the  rays,  after  impinging  on  the  metal,  will 
be  reflected  and  will  come  together  again  at  the  other  focus, 
which  will,  therefore,  be  brilliantly  illuminated.* 

The  same  is  true  of  heat,  since  heat  rays  are  reflected  from 
a  polished  surface  by  the  same  law  as  that  of  light  rays.  If, 
then,  a  candle  is  placed  at  one  focus  and  some  gunpowder  at  the 
other,  the  powder  can  be  ignited  by  the  heat  from  the  candle. 

Sound  waves  behave  in  a  similar  manner.  The  story  is  told 
of  the  Ratskeller  jn  Bremen,  the  walls  of  which  are  shaped 
somewhat  like  an  ellipse,  that  the  city 
fathers  were  remarkably  well  informed 
concerning  the  feelings  and  views  of  the 
populace.  For,  the  former  drank  their  FIG.  9 

wine  at  a  table  which  was  situated  at  a 

focus,  and  thus  could  hear  distinctly  the  conversation  at  a  dis- 
tant table,  which  stood  at  the  other  focus  and  about  which  the 
Burger  congregated. 


lope  and  Equation  of  the  Tangent.     The  student  will 
next  turn  to  Ch.  IX,  §  2,  where  the  slope  of  the  ellipse 

—  +  y~  =  1 
a2      ft"2 

*  The  statement  is,  of  course,  strictly  true  only  for  such  rays  as  travel 
in  the  plane  through  the  foci,  which  is  perpendicular  to  the  elements  of 
the  cylinder  formed  by  the  polished  band.  Since,  however,  only  a  nar- 
row strip  of  this  cylinder  is  used,  other  rays  will  pass  veiy  near  to  the 
second  focus  and  contribute  to  the  illumination  there. 


112 

at  the  point  (x1}  yi)  is  found  to  be 

j, 
(1) 


o 

The  equation  of  the  tangent  line  at  this  point  is 


(2) 


62 


Latus  Rectum.     The  latus  rectum  of  an  ellipse  is  defined  as 
a  chord  perpendicular  to  the  major  axis  and  passing  through 
a  focus.     The  term  is  also  used  to  mean 
the  length  of  such  a  chord. 
Thus,  in  the  ellipse 

25  + 16  =  1' 
FIG.  10 

one  focus  is  at  the  point  (3,  0).  The  length 
of  the  latus  rectum  is  twice  that  of  the  positive  ordinate 
corresponding  to  this  point.  Setting,  then,  x  =  3  in  the  equa- 
tion of  the  curve  and  solving  for  that  ordinate,  we  have 

y^_1 9_  _  16  _  16 _  Q, 

16         ~25~25'        y~~5~ 

Hence  the  length  of  the  latus  rectum  is  6£. 

EXERCISES 

1.  Find  the  equation  of  the  tangent  to  the  ellipse 

Ji-  +  #l  =  1 
225     25 

at  the  point  (9,  4).  Ans.   x  +  4i/  =  25. 

2.  Find  the  equation  of  the  normal  to  the  ellipse  of  Ex.  1 
at  the  same  point.  Ans.   4 a;  —  y  =  32. 

3.  At  what  point  does  the  tangent  to  the  ellipse 


at  the  point  (—1,2)  cut  the  axis  of  y  ? 


THE   ELLIPSE 


113 


4.  At  what  angle  does  the  straight  line  through  the  origin; 
which  bisects  the  angle  between  the  positive  axes  of  coordinates, 
cut  the  ellipse  3z2  +  4y2  =  7  ?  Ans.   81°  53'. 

5.  Find  the  area  of  the  triangle  cut  off  from  the  first  quad- 
rant by  the  tangent  to  the  ellipse  of  Ex.  3  at  the  point  (1,  2). 

Ans.   8|. 

6.  Find  the  length  of  the  latus  rectum  of  the  ellipse  of 
Ex.  1.  Ans.    3|. 

7.  The  same  for  the  ellipse  of  Ex.  3. 

8.  Show  that  the  length  of  the  latus  rectum  of  the  ellipse 


b<a, 


is  given  by  any  one  of  the  expressions 


-  e2 ;        2  o(l  -  e*). 

H 

Find  its  value  in  terms  of  c  and  e. 

9.   Find  the  length  of  the  latus  rectum  of  the  ellipse 

25  x2  +  16  y*  =  400.  ^4ns.   6|. 

10.    Prove  that  the  minor  axis  of  an  ellipse  is  a  mean  pro- 
portional between  the  major  axis  and  the  latus  rectum. 

7.  A  New  Locus  Problem.  Given  a  line  D  and  a  point  F 
distant  m  from  D.  To  find  the  locus  of  a  point  P  such  that 
the  ratio  of  its  distance  FP  from  F  to 
its  distance  MP  from  />  is  always 
equal  to  a  given  number,  c : 


(i) 

^  } 


MP 


or      FP=eMP. 


It  is  understood  that  P  shall  be  re- 
stricted to  the  plane  determined  by  F 
and  D. 

If,  in    particular,   e  =  1,    the    locus 


FIG.  11 


114  ANALYTIC   GEOMETRY 

is   a    parabola   with   D   as    directrix  and  F  as   focus  ;    Ch. 
VI,  §  1. 

To  treat  the  general  case,  let  D  be  taken  as  the  axis  of  y 
and  let  the  positive  axis  of  x  pass  through  F.     Then 


FP  =  V(z  —  m)2+  y\         MP  =  ±x, 

the  lower  sign  holding  only  when  x  is  negative,  and  (1)  be- 
comes 


(2)  V(as  -  m)2+  y2  =  ±  e x. 

On  squaring  and  transposing  we  obtain  the  equation : 

(3)  (1  -  £2)x2  -  2 mx  +  if  +  m2=  0. 

This  is  the  equation  of  the  proposed  locus. 
The  student  will  now  turn  to  Ch.  XI  and  study  carefully 
§1. 

EXERCISES 

1.  Take  c  =  \  and  m  =  3,  the  unit  of  length  being  1  cm. 
With  ruler  and   compasses   construct  a  generous  number  of 
points  of  the  locus,*  and  then  draw  in  the  locus  with  a  clean, 
firm  line. 

2.  Work  out   the  equation  of  the  locus  of  Ex.  1  directly, 
using  the  method  of  the  foregoing  text,  but  not  looking  at  the 
formulas.  Ans.   3  x*  +  4  y2  -  24  x  +  36  =  0. 

3.  Take  c  =  £  and  m  =  4,  the  unit  of  length  being  1  cm. 
Draw  the  locus  accurately,  as  in  Ex.  1. 

4.  Work  out  directly  the  equation  of  the  locus  of  Ex.  3. 

Ans.   16  x2  +  25  y2  -  200  x  =  -  400. 

5.  By  means  of  a  transformation  to  parallel  axes  show  that 
the  curve  of  Ex.  2  is  an  ellipse  whose  center  is  at  the  point 
(4,  0)  and  whose  axes  are  of  lengths  4  and  2V3.     What  is  its 
eccentricity  ? 

*  The  details  of  the  construction  are  an  obvious  modification  of  the 
corresponding  construction  for  the  parabola  in  Ch.  VI,  §  1.  A  circle  of 
arbitrary  radius  is  drawn  with  its  center  at  F,  and  this  circle  is  cut  by  a 
parallel  to  Z>,  whose  distance  from  D  is  twice  the  radius  of  the  circle. 


THE   ELLIPSE 


115 


6.    Show  that  the  curve  of  Ex.  4  is  an  ellipse  whose  axes 
are  7^  and  6.  >  What  is  its  eccentricity  ? 

8.   Discussion  of   the   Case  e  <  1.     The    Directrices.     From 

equation  (3)  of  §  7  follows: 

C\    .-.  .  .«>  -   .  o 

(1) 


1*>  "I  *>  "1  9 

—  (.-  1  —  £2  1  —  £2 

The  first  two  terms  on  the  left-hand  side  are  also  the  first  two 
in  the  expansion  of 

x -     m    Y=  z2        2m    x  4.       m* 
1-eV  1-e2         (1-e2)2 

If,  then,  we  add  the  third  term  of  the  last  expression  to  both 
sides  of  (1),  we  shall  have : 

o       2m  m2  y-  m2  m2 


1  —  e2 


or 

(2) 


_e2         (1_C2)2         1_ 


1  - 


1  —  e2       (1  —  c2)2 

This  equation  reminds  us  strongly  of  the  equation  of  an 
ellipse.  In  fact,  if  we  transform  to  parallel  axes  with  the 
new  origin,  0',  at  the  point 

m 
%o  —  ij 2  >         2/o  —  "> 

the  equations  of  transformation  are 

(3)  x'  =  x         m  y'  =  y, 
and  (2)  then  takes  on  the  form 

(4)  z'2  + 
or 

(5) 

where 

(6) 


1_C2          (1_£2)2' 


FIG.  12 


em 

1  -( 


i 

b  = 


VF^72 


116 


ANALYTIC   GEOMETRY 


Thus  the  locus  is  seen  to  be  an  ellipse  with  its  center,  0',  at 
the  point 

(7) 


the  semi-axes  being  given  by  (6). 

The  value  of  c  is  given  by  the  equation  c2  =  a2  —  62.     Hence 


(8) 


c  = 


1-e2 


The  eccentricity,  e  =  c/a,  is  now  seen  to  be  precisely  c : 


i.e.  the  given  constant,  e,  turns  out  to  be  the  eccentricity  of  the 
ellipse. 

Finally,  F  is  one  of  the  foci.     For,  the  distance  from  F  to  O' 
is 


OOf-OF  = 


m 


19  19' 

—  ej  1  —  e2 

and  this,  by  (8),  is  precisely  c. 

The  line  D  is  called  a  directrix  of  the  ellipse.     Its  distance 
from  the  center  is 

rm         m  me    1      a 

(J\J  —  • 


TTie  Directrices.     From   the   symmetry  of  the  ellipse  it  is 
clear  that  there  is  a  second  directrix,  Df,  on  the  other  side  of 

the  conjugate  axis,  parallel  to  that 
axis,  and  at  the  same  distance 
from  it  as  D.  This  line  D'  and 
the  focus  F'  stand  in  the  same  re- 
lation to  the  ellipse  as  the  first 
D'  line,  D,  and  the  focus  F.  Thus 
the  ellipse  is  the  locus  of  a  point 
so  moving  that  its  distance  from  a 

focus  always  bears   to  its   distance  from   the  corresponding 
directrix  the  same  ratio,  e,  the  eccentricity. 

Since  the  distance  of  D  from  the  center  of  the  ellipse  is  a/e, 
the  equations  of  the  directrices  of  the  ellipse 


FIG.  13 


THE    ELLIPSE  117 

^  +  5i=1'  a>6» 

are 

a  a 

e  e 

EXERCISES 

1.  Show  that  the  distances  of  the  vertices,  A  and  A',  from 
0  are: 

OA  =  —— ,        OA'  =  ^— 

2.  Collect  the  foregoing  results  in  a  syllabus,  arranged  in 
tabular  form,  giving  each  of  the  quantities  a,  6,  c,  00',  OA, 
OA',  OF,  OF'  in  terms  of  m  and  e. 

3.  Work  out  each  of  the  quantities  of  Ex.  2  directly  for  the 
ellipse  of  §  7,  Ex.  4,  and  verify  the  result  by  substituting  the 
values  e  =  |,  m  ==  4  in  the  formulas  of  the  syllabus. 

4.  Between  the  Jive  constants  of  the  ellipse,  a,  b,  c,  e,  m, 
there  exist  three  relations,  which  may  be  written  in  a  variety 
of  ways  ;  as,  for  example, 

'  "1  ^ 

i)  a"-  =  V-  +  c1 ;         ii)  e  =  - ;         iii)  m  —     ~e  a. 

a  e 

By  means  of  these  relations,  any  three  of  the  five  quantities 
can  be  expressed  in  terms  of  the  other  two.  Thus,  in  Ex.  2, 
m  and  e  are  chosen  as  the  quantities  in  terms  of  which  all 
others  shall  be  expressed. 

Taking  the  semi-axes,  a  and  6,  (a  >  6),  as  the  preferred  pair, 
express  the  other  quantities  in  terms  of  them. 

5.  Show  that  the  tangent  to  the  ellipse 

25  +  16  =  :l 

at  an  extremity  of  a  latus  rectum  cuts  the  transverse  axis  in 
the  same  point  in  which  this  axis  is  cut  by  a  directrix. 


118 


ANALYTIC   GEOMETRY 


6.  The  same  for  any  ellipse. 

7.  Prove  directly  that,  if  P  is  any  point  of  the  ellipse 

s+S-1- 

the  ratio  of  its  distance  from  a  focus  to  its  distance  from  the 
.^•-corresponding  directrix  is  equal  to  the  eccentricity. 

Show  that  in  an  ellipse  the  major  axis  is  a  mean  propor- 
tional between  the  distance  between  the  foci  and  the  distance 
between  the  directrices. 

9.  Show  that  the  distances  from  the  center  and  a  focus  of 
an  ellipse  to  the  directrix  corresponding  to  the  focus  are  in 
the  same  ratio  as  the  squares  of  the  semi-axis  major  and  the 
semi-axis  minor. 

9.  The  Parabola  as  the  Limit  of  Ellipses.  We  have  proved 
that,  when  e  <  1,  equation  (3),  §  7,  represents  an  ellipse  with 

eccentricity  e  =  e.  We 
know  that ,  if  e  =  1,  the 
equation  represents  a 
-  parabola.  If,  then,  in  the 
equation  we  allow  e  to 
approach  1  through  values 
<  1,  the  ellipse  which  the 
equation  defines  ap- 
proaches a  parabola  as  its 
limit. 

We  can  visualize  the 
ellipse,  going  over  into 
a  parabola,  by  drawing 
a  number  of  ellipses 
having  the  same  value  of 
are  increasing  toward  1 


FIG.  14 


which 


ra,  but  having  values  for 
as  their  limit,  viz.  e  =  -£,  e  =  f ,  e  =  -§-,  •••.  The  directrix  D, 
along  the  axis  of  y,  and  the  focus  F :  (ra,  0)  are  the  same  for 
all  the  ellipses.  But  the  center  0'  and  the  right-hand  vertex 


THE    ELLIPSE 


119 


A'  of  each  successive  ellipse  are  farther  away  from  0,  and 
their  distances  from  0,  namely, 

m  ^  A,        m 


00'  = 


OA'  = 


R 


1         •»»  i 

1  —  e2  1  —  e 

increase  without  limit.  Thus,  as  e  approaches  1,  the  ellipse 
approaches  as  its  limit  the  parabola  whose  directrix  is  D  and 
whose  focus  is  F. 

10.  New  Geometrical  Construction  for  the  Ellipse.  Para- 
metric Representation.  Let  it  be  required  to  draw  an  ellipse 
when  its  axes,  AA'  and  BB',  are  given. 
Describe  circles  of  radii  a  =  OA  and 
b  =  OB,  with  the  origin  0  as  the 
common  center.  Draw  any  ray  from 
0,  making  an  angle  <f>  with  the  posi- 
tive axis  of  x,  as  shown  in  the 
figure.  Through  the  points  Q  and  R 
draw  the  parallels  indicated.  Their 
point  of  intersection,  P,  will  lie  on 
the  ellipse.  For,  if  the  coordinates  of 
Pbe  denoted  by  (x,  y),  it  is  clear  that 

(1)  x  =  a  cos  <£,         y  =  b  sin  <£. 

From  these  equations  <£  can  be  eliminated  by  means  of  the 
trigonometric  identity 

sin2  <f>  -f  cos2  0  =  1. 
Hence 

(2)  *  +  £«!. 

a2     &2 

Conversely,  any  point  (x,  y}  on  the  ellipse  (2)  has  corre- 
sponding to  it  an  angle  0,  for  which  equations  (1)  are  true. 

Equations  (1)  afford  what  is  known  as  a  parametric  repre- 
sentation of  the  coordinates  of  a  variable  point  (#,  y)  of  the 
ellipse  in  terms  of  the  parameter  0.  When  b  =  a,  the  ellipse 
becomes  a  circle,  and  the  equations  (1)  become 

(3)  x  =  a  cos  0,        y  =  a  sin  0. 


FIG.  15 


120  ANALYTIC   GEOMETRY 

These  parametric  representations,  though  little  used  in  Ana- 
lytic Geometry,  are  an  important  aid  in  the  Calculus. 

The  larger  of  the  two  circles  in  Fig. 
15    is    commonly    called    the    auxiliary 
)P:(x,y)  circle  of  the  ellipse,  and  the  points  R 
and  P  are  known  as  corresponding  points. 
The  angle  <£  is  called  the  eccentric  angle. 


EXERCISE 

16  By  means  of  the  foregoing   method, 

draw  on  squared  paper  an  ellipse  whose 
axes  are  of  length  4  cm.  and  6  cm. 

EXERCISES  ON   CHAPTER  VII 

1.  The  earth  moves  about  the  sun  in  an  elliptic  orbit.*     The 
shortest  and  longest  distances  from  it  to  the  sun  are  in  the 
ratio  29  :  30.     What  is  the  eccentricity  of  the  orbit  ? 

2.  Show  that  the  slopes  of  the  tangents  to  an  ellipse  at  the 
extremities  of  the  latera  recta  are  ±  e. 

3.  The  axes  of  an  ellipse  which  goes  through  the  points 
(4,  1),  (2,  2)  are  the  axes  of  coordinates.     Find  its  equation. 

4.  The  center  of  an  ellipse  is  in  the  origin  and  the  foci  are 
on  the  axis  of  x.     The  ellipse  has  an  eccentricity  of  f  and  goes 
through  the  point  (12,  4).     What  is  its  equation? 

-r2      ?/2      1  fiQ 

A  <~        t       V  -LUc7 

A.US. h  --—  — ' 

25^16      25 

5.  Solve  the  preceding  problem  if  the  foci  may  lie  on  either 
axis  of  coordinates. 

6.  Find  the  equations  of  the  ellipses  which  have  the  axes 
of  coordinates  as  axes,  go  through  the  point  (3,  4),  and  have 
their  major  and  minor  axes  in  the  ratio  3  :  2. 

7.  Show  that  the  ellipses  represented  by  the  equation 


*  The  planets  describe  ellipses  about  the  sun  as  a  focus,  and  the  comets 
usually  describe  parabolas  with  the  sun  as  the  focus. 


THE   ELLIPSE  121 

where  c1  is  an  arbitrary  positive  constant,  are  similar.     What 
is  the  common  value  of  the  eccentricity  ? 

8.  How  many  ellipses  are  there  with  eccentricity  ^,  having 
their  centers  in  the  origin  and  their  foci  on  the  axis  of  a? 
Deduce  an  equation  which  represents  them  all. 

Ans.   3  a;2 +  4^  =  c2. 

9.  The  foci  of  an  ellipse  lie  midway  between  the  center 
and  the  vertices.     What  is  the  eccentricity  ?     How  many  such 
ellipses  are  there,  with  centers  in  the  origin  and  foci  on  the 
axis  of  x?     Write  an  equation  which  represents  them  all. 

10.  The  line  joining  the  left-hand  vertex  of  an  ellipse  with 
the  upper  extremity  of  the  minor  axis  is  parallel  to  the  line 
joining  the  center  with  the  upper  extremity  of  the  right- 
hand  latus  rectum.  Answer  the  questions  of  the  preceding 
"-''' 

foci  of  an  ellipse  subtend  a  right  angle  at  either 
extremity  of  the  minor  axis.  What  is  the  eccentricity? 
Find  the  equation  of  all  such  ellipses  with  centers  in  the 
origin  and  foci  on  the  axis  of  y. 

12.  Prove  that  the  ratio  of  the  distance  from  a  focus  of  an 
ellipse   to   the   intersection  with   the   transverse  axis  of   the 
normal  at  a  point  P,  and  the  distance  from  this  focus  to  P 
equals  the  eccentricity  of  the  ellipse. 

13.  The  projections  of  a  point  P  of  an  ellipse  on  the  trans- 
verse and  conjugate  axes  are  P^  and  P2.     The  tangent  at  P 
meets  these  axes  in  7\  and  T2.     Prove  that  OP}  •  OTl  =  a2  and 
OP?  -  OT2  =  b2,  where  0  is  the  center   and  a  and  b  are  the 
semi-axes  of  the  ellipse. 

14.  Prove  that  the  segment  of  a  tangent  to  an  ellipse  be- 
tween the  point  of  contact  and  a  directrix  subtends  a  right 
angle  at  the  corresponding  focus. 

15.  Determine  the  points  of  an  ellipse  at  which  the  tangents 
have  intercepts  on  the  axes  whose  absolute  values  are  propor- 
tional to  the  lengths  of  the  axes. 


122  ANALYTIC   GEOMETRY 

16.  Through  a  point  M  of  the  major  axis  of  an  ellipse  a  line 
is  drawn  parallel  to  the  conjugate  axis,  meeting  the  ellipse  in 
P  and  the  tangent  at  an  extremity  of  the  latus  rectum  in  Q. 
Show  that  the  distance  MQ  equals  the  distance  of  P  from  the 
focus  corresponding  to  the  latus  rectum  taken. 

17.  Prove  that  the  line  joining  a  point  P  of  an  ellipse  with 
the  center  and  the  line  through  a  focus  perpendicular  to  the 
tangent  at  P  meet  on  a  directrix. 

18.  Prove  that  the  distance  from  a  focus  F  to  a  point  P  of 
an  ellipse  equals  the  distance  from  F  to  the  tangent  to  the 
auxiliary  circle  at  the  point  corresponding  to  P. 

19.  Find  the  equation  of  a  circle  which  is  tangent  to  the 
ellipse 

*  +  £_! 

a?      &2  - 
at  both  ends  of  a  latus  rectum. 

20.  In  an  ellipse  whose  major  axis  is  twice  the  minor  axis, 
a  line  of  length  equal  to  the  minor  axis  has  one  end  on  the 
ellipse,  the  other  on  the  conjugate  axis.     The  two  ends  are 
always  on  opposite  sides  of  the  transverse  axis.     Prove  that 
the  mid-point  of  the  line  lies  always  on  the  transverse  axis. 

21.  A  number  of  ellipses  have  the  same  major  axis  both  in 
length  and  position.     A  tangent  is  drawn  to  each  ellipse  at 
the  upper  extremity  of  the  right-hand  latus  rectum.     Prove 
that  these  tangents  all  pass  through  a  point. 

Exercises  22-28.  In  these  exercises,  in  which  properties  in- 
volving an  arbitrary  point  P  of  an  ellipse  are  to  be  proved,  it 
will,  in  general,  be  necessary  to  make  actual  use  of  the  equa- 
tion expressing  the  fact  that  the  point  P  lies  on  the  ellipse. 

22.  The  tangent  to  an  ellipse  at  a  point  P  meets  the  tan- 
gent at  one  vertex  in  Q.     Prove  that  the  line  joining  the  other 
vertex  to  P  is  parallel  to  the  line  joining  the  center  to  Q. 

23.  The  lines  joining  the  extremities  of  the  minor  axis  with 
a  point  P  of  an  ellipse  meet  the  transverse  axis  in  the  points 


THE    ELLIPSE  123 

M  and  N.     Prove  that  the  semi-axis  major  is  a  mean  propor- 
tional between  the  distances  from  the  center  to  M  and  N. 

24.  Prove  the  theorem  of  the  preceding  exercise  when  the 
major  and  minor  axes,  and  the  transverse  and  conjugate  axes, 
are  interchanged. 

25.  Show    that   the   segment  of  a   directrix,  between   the 
points  of  intersection  of  the  lines  joining  the  vertices  with  a 
point  011  an  ellipse,  subtends  a  right  angle  at  the  correspond- 
ing focus. 

26.  Prove  that  the  product  of  the  distances  of  the  foci  of 
an  ellipse  from  a  tangent  is  a  constant,  independent  of  the 
choice  of  the  tangent. 

27.  Let  F'  and  F  be  the  foci  of  an  ellipse  and  P  any  point 
on  it.     Prove  that  62 :  FK-  =  F'P :  FP,  where  FK  is  the  dis- 
tance from  F  to  the  tangent  at  P. 

28.  The  normal  to  an  ellipse  at  a  point  P  meets  the  axes  in 
NI  and  N2.     Show  that  PN^  •  PN2  is  equal  to  the  product  of 
the  focal  radii  to  P. 

Loci 

29.  A  point  moves  so  that  the  product  of  the  slopes  of  the 
two  lines  joining  it  to  two  fixed  points  is  a  negative  constant. 
What  is  its  locus  ? 

30.  A  circle  whose  diameter  is  10  cm.  is  drawn,  center  at  0. 
On  a  radius  OA  a  point  B  is  marked  distant  4  cm.  from  0. 
If  OQ  is  any  second  radius,  show  how  to  construct,  with  ruler 
and  compasses,  a  point  P  on  OQ,  whose   distance  from  the 
circle  equals  its  distance  from  B.     In  this  way  plot  a  number 
of  points  on  the  locus  of  P. 

31.  Find  the  equation  of  the  locus  of  the  point  P  of  the 
preceding  exercise.     Take   the   origin   of   coordinates  at   the 
mid-point  of  OB. 

32.  The  base  of  a  triangle  is  fixed  and  the  product  of  the 
tangents  of  the  base  angles  is  a  positive  constant.     Find  the 
locus  of  the  vertex. 


CHAPTER   VIII 


FIG.  1 


THE   HYPERBOLA 

1.  Definition.  A  hyperbola  is  defined  as  the  locus  of  a  point 
P,  the  difference  of  whose  distances  from  two  given  points,  F 

and  F',  is  constant.  It  is  found 
convenient  to  denote  this  constant 
by  2  a.  Then 

FP-  F'P=2a, 
or  F'P-FP=2a. 

It  is  understood,  of  course,  that  P 
is  restricted  to  a  particular  plane 
through  F  and  F'. 

The  points  F  and  F'  are  called  the  foti  of  the  hyperbola. 
It  is  clear  that  2  a  must  be  less  than  the  distance  between 
them.  Denote  this  distance  by  2  c. 

Geometrical  Construction.  Draw  the  indefinite  line  FFr, 
mark  the  mid-point,  0,  of  the  segment  FF',  and  the  points  A 

and    A'    each    at    a    distance    a 

P  P 

from  0: 

<X4=(M'  =  a;        OF=OF'  =  c. 

— h — 1 
The  point  ^4.  lies  on  the  locus ;       F    * 

for,  v^ 

J"-4  =  c  +  a, 


A     F 


FA  =  c-  a, 
and  hence 


FIG.  2 


—  #4  =  2  a. 


Likewise,  A'  lies  on  the  curve. 

Mark  any  point,  N,  to  the  right  of  F.     With  radius  AN  and 
center  JF1,  describe  a  circle.     Next,  with  radius  ^.'^Tand  center 

124 


THE    HYPERBOLA 


125 


F',  describe  a  second  circle.  The  points  P  and  Q  in  which 
these  circles  intersect  are  points  of  the  locus.  For, 

F'P-  FP=  A'N-AN=  A' A  =  2  a. 

Two  more  points,  P'  and  Q',  can  be  obtained  from  the  same 
pair  of  settings  by  interchanging  the  centers,  F  and  F',  of  the 
circles. 

By  repeating  the  construction  a  number  of  times,  a  goodly 
array  of  points  of  the  hyperbola  can  be  obtained.  These 
points  will  lie  on  two  distinct  arcs, 
symmetric  to  each  other  in  the 
perpendicular  bisector  BOB'  of 
FF'.  Thus  it  will  be  seen  that 
the  hyperbola  consists  of  two 
parts,  or  brandies,  as  they  are 
called.  These  branches,  besides 
being  the  images  of  each  other  in 
BB',  are  each  the  image  of  itself 

in  FF'.  It  is  natural  to  speak  of  the  indefinite  straight  lines 
FF'  and  BB'  as  the  axes  of  the  hyperbola.  FF'  is  called  the 
transverse,  BB'  the  conjugate  axis ;  0  is  the  center,  and  A,  A' 
are  the  vertices. 

EXERCISES 

~l]  Taking  c  =  3  cm.  and  a  =  2.  cm.,  make  a  clean  drawing  of 
the  corresponding  hyperbola. 

2.  Reproduce  the  drawing  on  a  rec- 
tangular card  and,  with  a  sharp  knife 
or  a  small  pair  of  scissors,  cut  out  the 
center  of  the  card  along  the  hyperbola 
and  two  parallels  to  the  transverse  axis. 
On  the  templet  which  remains  make 
holes  at  the  foci  and  draw  the  two  axes. 


FIG.  3 


FIG.  4 


2.   Equation  of  the  Hyperbola.     The  treatment  here  is  paral- 
lel to  that  of  the  ellipse,  Ch.  VII,  §  3.     Let  the  transverse  axis 


ANALYTIC  GEOMETRY 


be  chosen  as  the  axis  of  x ;  the  conjugate  axis,  as  the  axis  of  y. 
Then  the  equation  of  the  right-hand  branch  of  the  hyperbola 
can  be  written  in  the  form 

(P:(x,y)       ,„ .         n • — 


F:C,O) 


F:(-c,o) 


FIG.    5 


-  V(a?  -  c)2  +  y*  =  2  a. 
Transpose  the  first  radical  and  square  : 


Tj 

Hence 


—  4  aV(x  +  c)2+  y2  -f  4  a2. 


-f  c)2  +  y2  =  a2  +  ex. 


(2)  a 
Square  again  : 

a2a?  +  2  a2ca;  +  a2c2  -f  ay  =  a4  +  2  a?cx 
or 

(3)  (a2  —  C2)a2  +  a7y2=«2(a2  —  c2). 

This  is  precisely  the  same  equation  that  presented  itself  in 
the  case  of  the  ellipse  ;  but  the  locus  is  a  curve  of  wholly  dif- 
ferent nature.  The  reason  is,  that  a  and  c  have  different 
relative  values.  In  the  ellipse,  a  was  greater  than  c,  and  hence 
a-  —  c-  was  positive.  It  could  be  denoted  by  62.  Here,  a  is 
less  than  c  ;  a2  —  c-  is  negative,  and  it  cannot  be  set  equal  to  62.. 
It  can,  however,  be  set  equal  to  —  62.  This  we  will  do  : 

(4)  a2  —  c2  =  —  62,         or        c2  =  a2  +  ft2, 

thus  defining  the  quantity  b  in  the  case  of  the  hyperbola  by 
the  equation  : 

6=Vc2  —  a2. 

The  final  equation  between  x  and  y  can  now  be  written  in 
the  form 

T-  4/2 

(5)  ^_  _£-=!. 

a2     &2 

This  equation  is  satisfied  by  the  coordinates  of  all  points  on 
the  right-hand  branch,  as  is  seen  from  the  way  in  which  it 
was  deduced.  It  is,  however,  also  satisfied  by  the  coordinates 
of  all  points  on  the  left-hand  branch.  For  such  a  point,  the 


THE   HYPERBOLA  127 

signs  of  both  radicals  in  (1)  will  be  reversed.  Starting,  now, 
with  the  new  equation  and  proceeding  as  before,  we  find  the 
same  equation  (3),  which  we  may  again  write  in  the  form  (5), 
and  thus  the  truth  of  the  statement  is  established. 

Is  (5)  satisfied  by  the  coordinates  of  still  other  points  ?  To 
answer  this  question,  let  (x,  y)  be  any  point  whose  coordinates 
satisfy  (5).  Then,  starting  from  (5),  we  retrace  our  steps, 
admitting,  each  time  that  we  extract  a  square  root,  both  signs 
of  the  radical  as  conceivably  possible.  Thus  we  can  be  sure 
that  (x,  y)  will  satisfy  one  of  the  four  equations 


±  V(»  +  c)-+  f-±  V(*  -c)2  +  f-  =  2a, 

corresponding  to  the  four  conceivable  choices  of  the  signs  of 
the  radicals : 

i)  +  5  iii)  -  5 

ii)         +      -5  iv)         +     +. 

If  (x,  y)  satisfies  i)  or  ii),  the  point  lies  on  the  hyperbola. 
The  other  two  cases  are  impossible.  For,  case  iii)  says  that  a 
negative  quantity  is  equal  to  a  positive  quantity,  and  case  iv) 
says  that  F'P  +  FP  =  2 a.  Now  F'P  +  FP,  being  the  sum  of 
two  sides  of  the  triangle  FPF',  is  greater  than  the  third  side, 
FF',  or  2  c.  But  2  a  is  actually  less  than  2  c.  Hence  we  have 
a  contradiction,  and  this  case  cannot  arise. 

We  have  shown  then,  finally,  that  (5)  is  the  equation  of  the 
hyperbola. 

EXERCISE 
Plot  the  hyperbola 

^!_^!_=i 
25     16 

directly  from  its  equation,  taking  1  cm.  as  the  unit  of  length. 

3.  Axes,  Eccentricity,  Focal  Radii.  The  transverse  and  the 
conjugate  axis  have  already  been  defined  in  §  1.  The  segment 
AA'  of  the  transverse  axis  is  called  the  major  axis,  and  this 
term  is  also  applied  to  its  length,  2  a.  The  segment  BB'  of 
the  conjugate  axis,  whose  center  is  at  0  and  whose  length  is 


128  ANALYTIC   GEOMETRY 

26,  is  called  the  minor  axis,  and  this  term  is  also  applied  to 

its  length,  2  6. 

The  major  axis  of  an  ellipse  is  always  longer  than  the  minor 

axis.  In  the  case  of  the  hyperbola,  however,  this  is  not  al- 
ways true.  For  example,  if  2c  and  2  a 
are  taken  as  10  and  6  respectively,  then 
26  =  8.  Thus  the  major  axis  of  the 
hyperbola  is  to  be  understood  as  the 

principal  axis,  but  not  necessarily  as  the 
FIG.  6  ^  . 

longer  axis. 

The  eccentricity  of  the  hyperbola  is  defined  as  the  number 


Since  c  is  greater  than  a,  the  eccentricity  of  a  hyperbola  is 
always  greater  than  unity. 

The  eccentricity  characterizes  the  shape  of  the  hyperbola. 
All  hyperbolas  having  the  same  eccentricity  are  similar,  differ- 
ing only  in  the  scale  to  which  they  are  drawn,  and  conversely  ; 
cf.  Exercise  8. 

The  focal  radii  FP,  F'P  can  be  represented  by  simple  ex- 
pressions, similar  to  those  which  presented  themselves  in  the 
case  of  the  ellipse.  On  dividing  equation  (2),  §  2,  through  by 
a,  we  have : 

V(#  +  c)l  +  yl  =  a  +  ex. 

Hence,  when  P  is  a  point  of  the  right-hand  branch, 

(1)  F'P  =  ex  +  a. 
The  evaluation, 

(2)  FP  =  ex-  a, 

is  obtained  in  a  similar  manner.* 

If  P  is  a  point  of  the  left-hand  branch,  these  formulas 
become : 

(3)  F'P  =  -(ex  +  a);        FP=-  (ex  -  a). 

*  P  being  a  point  of  the  right-hand  branch,  x  is  positive  and  greater 
than  or  equal  to  a  ;  also,  e>  1.  Hence  ez>  a,  and  ex  —  a  is  positive, 
as  it  should  be. 


THE   HYPERBOLA  129 

EXERCISES 

Q  Find  the  lengths  of  the  axes,  the  coordinates  of  the  foci, 
and  the  value  of  the  eccentricity  for  each  of  the  following 
hyperbolas. 


Ana.    8,6;     (5,  0),  (-  5,  0)  ;     If 

(b\  x1--  y*-  =  a2.     Ana.  2  a,  2  a  ;    (a  V2,  0),  (-  a  V2,  0)  ;    V2. 
-  3?/2  =  24.  (e)    5tf  -6^  =  8. 

-  -  ^  =  4.  (/)  6  &  -  9y-  =  4. 

2.    If  the  eccentricity  of  a  hyperbola  is  2  and  its  major  axis 
is  3,  what  is  the  length  of  its  minor  axis  ?  Ans.   3  V3. 

[3^  How  far  apart  are  the  foci  of  the  hyperbola  in  Ex.  2  ? 

Ans.   6. 

4.    What  is  the  equation  of  the  hyperbola  whose  eccentricity 
is  V2  and  whose  foci  are  distant  4  from  each  other  ? 

Ans.   x1  —  y2  =  2. 

^    The  extremities  of  the  minor  axis  of  a  hyperbola  are  in 
the  points  (0,  ±  3)  and  the  eccentricity  is  2.     Find  the  equa- 
tion of  the  hyperbola. 
Tel   Show  that,  in  terms  of  a  and  b,  e  has  the  value 


m  Express  6  in  terms  of  a  and  e. 

8.  Prove  that  two  hyperbolas  which  have  the  same  eccen- 
tricity are  similar,  and  conversely. 

9.  Establish  formulas  (3). 

4.  The  Asymptotes.  Two  lines,  called  the  asymptotes,  stand 
in  a  peculiar  and  important  relation  to  the  hyperbola.  They 
are  the  lines 

^         and         ,«-*». 

a  a 


130 


ANALYTIC   GEOMETRY 


Let  a  point  P :  (x,  y)  move  off 
along  a  branch  of  the  hyperbola 


(1) 


a2      &2 


FIG.  7 


and  let  this  take  place,  for  def- 
initeness,  in  the  first  quadrant. 
The  slope  of  the  line  OP  is 


OM     x 
Since  the  coordinates  (x,  y)  of  P  satisfy  (1),  it  follows  that 

(2) 

and  hence 


(3) 


x     a 


When  P  recedes  indefinitely,  x  increases  without  limit,  and 
the  right-hand  side  of  this  equation  approaches  the  limit  b/a. 
Thus  we  see  that  the  slope  of  OP  approaches  that  of  the  line  OQ, 


as  its  limit,  always  remaining,  however,  less  than  the  latter 
slope,  so  that  P  is  always  below  OQ. 

It  seems  likely  that  P  will  come  indefinitely  near  to  this 
line  ;  but  this  fact  does  not  follow  from  the 
foregoing,    since   P  might   approach   a    line 
parallel  to  (4)  and  lying  below  it.     In  that 
case,  all  that  has  been  said  would  still  be  true. 

That  P  does,  however,  actually  approach 
(4)  can  be  shown  by  proving  that  the  dis- 
tance PQ  approaches  0  as  its  limit.  Now,  FIG.  8 


and,  from  (4), 


THE   HYPERBOLA  131 

Furthermore,  MP  is  the  y-coordinate  of  the  point  P  on  the 
hyperbola : 


Hence  PQ  =  -  [x  -  Vas4  -  a2]. 

a 

To  find  the  limit  approached  by  the  square  bracket,  we  re- 
sort to  an  algebraic  device.  The  value  of  the  bracket  will 
clearly  not  be  changed  if  we  multiply  and  divide  it  by  the 
expression  x  +  V#2  —  a2 : 

/-= »      (x  —  Va2  —  a2)  (x  +  Va;2  -  a2) 

x  —  V  #2  —  a2  =  •> — *•  • 

z  +  Va;2—  a2 

But  the  numerator  of  the  last  expression  reduces  at  once  to  a2. 
Hence 

x  — 


From  this  form  it  is  evident  that  the  bracket  approaches  0 
when  x  increases  indefinitely;  and  hence  the  limit  of  PQ  is 
zero,*  q.  e.  d. 

Similar  reasoning,  or  considerations  of  symmetry,  applied  in 
the  other  quadrants,  show  that  in  the  second  and  fourth 
quadrants  P  approaches  the  line 

(5)  y  =  -lx> 

while  in  the  third  quadrant,  as  in  the  first,  P  approaches  (4). 
The  equations  (4)  and  (5),  of  the  asymptotes,  can  also  be 
written  in  the  form 

SL-f^O,        *  +  Z  =  0. 

a     b  a     b 

*  The  limit  approached  by  the  variable  x  —  -\/&  —  a2  can  be  found 
geometrically  as  follows.      Construct  a  variable 
right  triangle,  one  leg  of  which  is   fixed  and  of 
length  a,  the  hypothenuse  being  variable  and  of  -Jxz-az 

length  x.    Then  the  above  variable,  x  —  Vx2  —  a2,  FIQt  9 

is  equal  to  the  difference  in  length  between  the 

hypothenuse  and  the  variable  leg.  This  difference  obviously  approaches 
0  as  x  increases  indefinitely. 


132 


ANALYTIC   GEOMETRY 


It  is  easy  to  remember  these  equations,  since  they  can  be 
written  down  by  replacing  the  right-hand  side  of  (1)  by  0,  fac- 
toring the  left-hand  side  : 


and  putting  the  individual  factors  equal  to  zero. 

The  slopes  of  the  asymptotes  are  b/a  and  —  b/a.  Conse- 
quently, the  asymptotes  make  equal  angles  with  the  transverse 
axis. 

Since  the  ratio  of  6  to  a  is  unrestricted,  the  asymptotes  can 
make  any  arbitrarily  assigned  angle  with  each  other.  If,  in 
particular,  b  =  a,  this  angle  is  a  right  angle,  and  the  curve  is 
called  a  rectangular,  or  equilateral,  hyperbola.  Its  equation  can 
be  written  in  the  form  : 
(6)  x"-  —  y1  =  a2. 

Its  eccentricity  is  e  =  V2. 

Construction  of  the  Asymptotes.     Mark  with  heavy  lines  the 

major  and  minor  axes,  and  through  ..the  extremities  of  each 

draw  Jines  parallel  to  the  other, 
thus  obtaining  a  rectangle.  The 
olagonals  of  this  rectangle,  pro- 
duced, are  the  asymptotes,  since 
their  slopes  are  clearly  ±  b/a. 

The  diagonals  of  the  rectangle 
have  lengths  equal  to  the  distance 
2  c  between  the  foci,  for,  c2  =  a?  +  62 

and  the  lengths  of  the  sides  of  the  rectangle  are  2  a  and  26. 

If  the  acute  angle  between  an  asymptote  and  the  transverse 

axis  is  denoted  by  a,  then 

e  =  sec  a. 


FIG.  10 


EXERCISES 


1.    Find  the  equations  and  slopes  of  the  asymptotes  of  the 
hyperbolas  of  Exercise  1,  §  3.     Draw  the  hyperbolas. 


HYPERBOLA  133 

Show  that  the  asymptotes  of  the  hyperbola 
Ax"-  -  By-  =  C, 

where  A,  B,  and  C  are  any  "three  positive  quantities,  are  given 
by  the  equations 

~\/Ax  +'V-B?/  =  0,         ~\/~Ax  —  ~\/By  =  0. 

|3\  Find  the  equation  of  the  hyperbola  whose  asymptotes 
make  angles  of  60°  with  the  axis  of  x  and  whose  vertices  are 
situated  at  the  points  (1,  0),  and  (—  1,  0).  Ans.  Sx"1  —  yz  =  3. 

>^I  Show  that  the  slopes  of  the  asymptotes  are  given  by 
the  expression  ±  Ve2  —  1. 

^  The  slope  of  one  asymptote  of  a  hyperbola  is  f .  Find 
the  eccentricity.  Ans.  e  -—  1^. 

6.  The  distance  of  a  focus  of  a  certain  hyperbola  from  the 
center  is  10  cm.,  and  the  distance  of  a  vertex  from  the  focus  is 
2  cm.     What  angle  do  the  asymptotes  make  with  the  conju- 
gate axis  ?  Ans.   53°  8'. 

7.  Show  that  the  circle  circumscribed  about  the  rectangle 
of  the  text  passes  through  the  foci. 

^8.^  A  perpendicular  di-opped  from  a  focus  F  on  an  asym- 
plote  meets  the  latter  at  E.  Show  that  OE  =  a,  and  EF  =  b. 

9.    Find  the  equation  of  the  equilateral  hyperbola  whose 
foci  are  at  unit  distance  from  the  center. 
[lOj  Find  the  equation  of  the  equilateral  hyperbola  which 
js  through  the  point  (—5,  4). 


Tangents.  The  method  of  finding  the  slope  of  an  ellipse, 
Ch.  IX,  §  2,  can  be  applied  to  the  hyperbola,  and  it  is  thus 
shown  that  the  slope  of  this  curve, 

l~f~2  =  1' 
at  the  point  (xl}  y^)  is 


134 


ANALYTIC  GEOMETRY 


The  equation  of  the  tangent  of  the  hyperbola  at  this  point  is 


(1) 


&2 


THEOREM.     The  tangent  of  a  hyperbola  at  any  point  bisects  the 
angle  between  the  focal  radii. 

To  prove  this  proposition  we  recall  the  theorem  of  Plane 
Geometry  which  says  that  the  bisector  of  an  angle  of  a  triangle 

divides  the  opposite  side  into  seg- 
ments which  are  proportional  to  the 
adjacent  sides.  It  is  easily  seen 
that  the  converse*  of  this  proposi- 
tion is  also  true,  and  hence  it  is 
sufficient  for  our  proof  to  show  that 


FlG 


FM 

We    already     have     simple     ex- 
pressions   for   the   numerators.     If 

P'-  (XD  y\)  be  a  point  of  the  right-hand  branch  of  the  curve, 
then,  by  §  3, 

FP  =  exl  —  a;        FP=  e^  +  a. 


To  compute  the  denominators,  find  where  the  tangent  at  P, 
whose  equation  is  given  by  (1),  cuts  the  axis  of  x.  Denoting 
the  abscissa  of  M  by  x',  we  have  : 


x  =  — 
xl 


Now, 

and 

But  c  =  ae,  and  so 

Thus 


FM=  OF  -  OM=  c  -  x', 


xl 


cxi  —  a2  =  a(exl—  a). 
c  —  x'  =  —  (exi  —  a), 


*  Let  the  student  prove  this  proposition  as  an  exercise. 


THE    HYPERBOLA  135 

and  we  arrive  finally  at  the  desired  expression  for  FM  : 


xl 
In  a  similar  manner  it  is  shown  that 

F'M=-(ex1  +  a). 
«i 

From  these  evaluations  it  appears  that 

FP__Xi  F'P      xv 

FM     a  F'M~  a  ' 

Hence  (2)  is  a  true  equation,  and  the  proof  is  complete  for  the 
case  that  P  lies  on  the  right-hand  branch.  Since,  however, 
the  curve  is  symmetric  in  the  conjugate  axis,  the  theorem  is 
true  for  the  left-hand  branch  also. 

Latus  Rectum.  The  latus  rectum  of  a  hyperbola  is  defined 
as  a  chord  passing  through  a  focus  and  perpendicular  to  the 
transverse  axis.  The  term  is  also  applied  to  the  length  of 
such  a  chord. 

EXERCISES 

|^J  Find  the  slope  of  the  hyperbola  4  x2  —  y*  =  15  at  the 
point  (2,  -  1).  Ans.  -  8. 

2.  Find  the  equation  of  the  tangent  of  the  hyperbola  of 
Ex.  1  at  the  point  there  mentioned.  .4ns.  8  x  +  y  =  15. 

J3j  Find  the  angle  at  which  the  line  through  the  origin  bi- 
secting the  angle  between  the  positive  axes  of  coordinates  cuts 
the  hyperbola  of  Ex.  1.  Ans.  30°  58'. 

4.    Find  the  length  of  the  latus  rectum  of  the  hyperbola 

rfi  »/2 

fg-|=l.  Ans.    4J. 

te^  Find  the  length  of  the  latus  rectum  of  the  hyperbola  of 
Ex.  1.  Ans.  15.49. 

6.   Find  the  equation  of  the  normal  of  the  hyperbola 

^._J/L=1 
25     144 


136 


ANALYTIC   GEOMETRY 


at  the  extremity  of  the  latus  rectum  which  lies  in  the  first 
juadrant.  Ans.   25  x  +  65  y  —  2197. 

Show  that  the  length  of  the  latus  rectum  of  the  hyperbola 


is 


2&2 


a2     62 


8.  Prove  that  the  tangents  at  the  extremities  of  the  latera 
recta  have  slopes  ±  e. 

/£l  In  an  ellipse,  the  focal  radii  make  equal  angles  with  the 
tangent.  Prove  this  theorem  by  the  method  employed  in  this 
paragraph  to  prove  the  corresponding  theorem  relating  to  the 
hyperbola. 

6.  New  Definition.  The  Directrices.  The  locus  defined 
in  Ch.  VII,  §  7,  can  now  be  shown  to  be  a  hyperbola  when 
c  >  1.  The  analytic  treatment  given  there  and  in  §  8  down 
to  equation  (2)  and  the  transformation  (3)  holds  unaltered 
for  the  present  case. 

When,  however,  c  >  1,  the  new  origin,  O7,  lies  to  the  left  of 

/  m  \ 

0,  in  the  point  ( — ,  0  ),  and  it  is  more  natural  to  write 

V     c2  - 1      J 
(3)  in  the  form 

(1)  ^  =  3+™ 


y  =y, 


and  likewise  (4)  as 
(2) 


x  *  — 


_  I         (£2  _ 


This  equation  passes  over  into  the  form 

(3)  *~'2 
on  setting 

(4)  a  = 


62 


FIG.  12 


e'-l' 


THE    HYPERBOLA  137 

Thus  the  locus  is  seen  to  be  a  hyperbola  with  its  center,  0', 

at  the  point  [  --  ,  0),  the  semi-axes  being  given  by  (4). 

\     «2  -  1     / 
The  value  of  c  is  given  by  the  equation  c2  =  cC-  +  62.     Hence 

(5) 


e2-! 

The  eccentricity,  e  =c/a,  is  seen  to  .be  precisely  c  : 

e  =  e, 

and  thus  the  given  constant,  e,  turns  out  to  be  the  eccentricity  of  the 
hyperbola. 

Finally,  F  is  one  of  the  foci.     For,  the  distance  from  0'  to  F 

is 

0'0+  OF  =  ^  * 


2      ?' 

£2  —  1 

and  this,  by  (5),  is  precisely  c. 

The  line  D  is  called  a  directrix  of  the  hyperbola.     Its  dis- 
tance from  the  center  is 


cm 


E"  —  L       f. 


The  Directrices.  There  is  a  second  directrix,  namely,  the 
line  D'  symmetric  to  D  in  the  conjugate  axis.  It  is  clear 
from  the  symmetry  of  the  figure  that  what  is  true  of  the  hy- 
perbola with  respect  to  the  focus  F  and  the  corresponding 
directrix  D  is  equally  true  with  respect  to  the  focus  F'  and 
the  directrix  D'.  Accordingly,  the  hyperbola  is  the  locus  of  a 
point  whose  distance  from  a  focus  bears  to  its  distance  from 
the  corresponding  directrix  a  fixed  ratio,  the  eccentricity'. 

The  equations  of  the  directrices  of  the  hyperbola, 


a2     &2 
are  x  =  -  and 


138  ANALYTIC   GEOMETRY 

EXERCISES 

£y[  Take  c  =  2  and  m  =  3,  the  unit  of  length  being  1  cm. 
With  ruler  and  compasses  construct  a  generous  number  of 
points  of  the  locus,  and  then  draw  in  the  locus  with  a  clean, 
firm  line.* 

2.  Work  out  the  equation  of  the  locus  of  Ex.  1  directly, 
using  the  method  of  Ch.  VII,  §  7,  but  not  looking  at  the 
formulas.  Ans.  3  x1  —  y1  +  6  x  =  9. 

/3J  By  means  of  a  transformation  to  parallel  axes  show 
that  the  curve  of  Ex.  2  is  a  hyperbola  whose  center  is  at  the 
point  (—  1,  0)  and  whose  axes  are  of  lengths  4  and  4V3. 

4.  Show  that  in  the  general  case  the  distances  of  the 
vertices,  A  and  A',  from  0  are : 


^5.1  Collect  the  results  of  this  paragraph  in  a  syllabus, 
arranged  in  tabular  form,  giving  each  of  the  quantities,  a,  6,  c, 
O'O,  OA,  A'O,  OF,  and  F'O,  in  terms  of  m  and  e. 

6.  Work  out  each  of  the  quantities  of  Ex.  5  directly  for 
the  curve  of  Ex.  2  and  verify  the  result  by  substituting  the 
values  e  =  2,  m  =  3  in  the  formulas  of  the  syllabus. 

rn    Show  that  the  tangent  to  the  hyperbola 

*l_2?=i 

16      9 

at  an  extremity  of  a  latus  rectum  cuts  the  transverse  axis  in 
the  same  point  in  which  this  axis  is  cut  by  a  directrix. 
8.    The  same  for  any  hyperbola. 

*  The  footnote  of  p.  114  applies  in  the  present  case  with  the  obvious 
modification  that  the  distance  of  the  parallel  from  D  must  now  be  half 
the  radius  of  the  circle.  Moreover,  two  parallels  to  D  must  now  be  drawn, 
the  second  one,  as  soon  as  the  radius  has  increased  sufficiently,  giving 
points  on  the  left-hand  branch. 


THE    HYPERBOLA  139 

(  91  Prove  directly  that,  if  P  is  any  point  of  the  hyperbola 

a?~V=    ' 

the  ratio  of  its  distance  from  a  focus  to  its  distance  from  the 
corresponding  directrix  equals  the  eccentricity. 

10.  Prove  that  the  ratio  of  the  distance  between  the  foci  of 
a  hyperbola  to  the  distance  between  the  directrices  equals  the 
square  of  theegcentricity. 


The  Parabola  as  the  Limit  of  Hyperbolas.     Summary. 

Equation  (3)  of  Ch.  VII,  §  7,  namely, 

(1)  (1  -  e2)z2  +  y1  -  2  mx  +  m"-  =  0, 

represents  a  hyperbola  when  c  >  1  and  a  parabola  when  «  =  1. 
If,  then,  we  let  c  approach  1  through  values  greater  than  1, 
the  hyperbola  which  (1)  represents  will  approach  a  parabola 
as  its  limiting  position. 

Suppose,  for  example,  that  we  take  m  =  2  and  let  e  take  on 
successively  the  values  2,  1^,  1-J,  li,  — .  Drawing  the  corre- 
sponding hyperbolas,  we  find  that,  whereas  the  directrix  D 
and  the  right-hand  focus  F  are  always  fixed,  the  center  and  the 
left-hand  vertex  keep  receding  to  the  left,  and  that  their 
distances  from  0,  namely, 

O'O—      m  A'O  —     m 

increase  without  limit.  Thus,  when  e  approaches  1,  the  left- 
hand  branch  of  the  hyperbola  recedes  indefinitely  to  the  left 
and  disappears  in  the  limit,  whereas,  meanwhile,  the  right- 
hand  branch  gradually  changes  shape  and  in  the  limit  becomes 
the  parabola  whose  directrix  is  D  and  whose  focus  is  F. 

Summary.  Let  us  now  combine  the  results  of  §  6  with  those 
of  §  8,  Ch.  VII.  We  have  proved  that  equation  (1)  repre- 
sents an  ellipse,  a  parabola,  or  a  hyperbola,  according  as  c<  1, 
e  =  1,  or  e  >  1.  In  case  of  the  ellipse  and  the  hyperbola  the 


140  ANALYTIC   GEOMETRY 

constant  e  turned  out  to  be  the  eccentricity  e.  We  are  led 
then  to  give  to  the  parabola  an  eccentricity,  namely, 
e  =  e  =  1. 

THEOREM.  The  locus  of  a  point  which  moves  so  that  its  dis- 
tance from  a  fixed  point  bears  to  its  distance  from  a  fixed  line, 
not  passing  through  the  fixed  point,  a  given  ratio  e  is  an  ellipse, 
a  parabola,  or  a  hyperbola,  according  as  e  is  less  than,  equal  to, 
or  greater  than  unity.  In  every  case  the  constant  e  equals  the 
eccentricity. 

Since  always  e  =  e,  we  may  suppress  e  in  future  work,  and 
use  e  exclusively.  Thus  equation  (1)  becomes 

(2)  (1  —  e'X  +  f'  —  2  mx  +  m2  i  =  0- 

The  theorem  furnishes  a  blanket  definition  for  the  ellipse, 
parabola,  and  hyperbola,  which  might  have  been  used  instead 
of  the  separate  definitions  which  we  have  given.  It  should 
be  noted,  however,  that  this  blanket  definition  does  not  include 
the  circle.  For,  if  we  set  e  =  0  in  (2),  the  equation  reduces  to 


which  represents  merely  the  focus  F  :  (m,  0). 

The  fact  that  the  blanket  definition  does  not  yield  a  circle 
as  a  special  case  in  no  way  discredits  the  circle  as  the  limiting 
form  of  an  ellipse  when  the  eccentricity  approaches  zero, 
Ch.  VII,  §  1.  The  reason  that  a  circle  cannot  be  defined  in 
the  new  manner  is  because  it  has  no  directrices.  When  the 
eccentricity  of  an  ellipse  approaches  zero,  the  major  axis 
remaining  constant,  the  distance  a/e  of  the  directrices  from 
the  center  increases  indefinitely,  so  that  in  the  limit,  when  the 
ellipse  becomes  a  circle,  the  directrices  have  disappeared.* 

*  It  is,  of  course,  possible  to  obtain  the  circle  as  a  limiting  curve  ap- 
proached by  ellipses  defined  in  the  new  way.  If  the  points  F  and  A  of 
Fig.  12,  Ch.  VII,  are  held  fast  and  m  is  allowed  to  increase  indefinitely, 
then  it  can  be  shown  that  e  approaches  zero  and  that  a  and  6  both  approach 
the  fixed  distance  AF.  Thus  the  variable  ellipse  approaches  a  circle  as 
its  limit. 


THE    HYPERBOLA 


141 


8.  Hyperbolas  with  Foci  on  the  Axis  of  y.  Conjugate 
Hyperbolas.  Let  the  student  show  that  the  equation  of  the 
hyperbola  whose  foci  are  at  the  points  (0,  ±  (7)  on  the  axis 
of  y  and  the  difference  of  whose  focal  radii  is  2J3  is 


where 


C2  = 


The  transverse  axis  of  this  hyperbola  is  the  axis  of  y ;  the 
conjugate  axis,  the  axis  of  x.  The  length  of  the  major  axis  is 
2B ;  that  of  the  minor  axis,  2  A.  The  eccentricity  is  C/B  and 
the  asymptotes  have  the  equations, 

—  =  0        and         — I-  —  =  0. 

,  A     B  A     B 

Conjugate  Hyperbolas.     The  two  hyperbolas, 


a2      62 


and 


have  the  same  asymptotes.  The  transverse  axis  of  each  is  the 
conjugate  axis  of  the  other,  and  the  major  axis  of  each  is  the 
minor  axis  of  the  other. 

Taken  together,  the  two 
hyperbolas  form  what  is 
called  a  pair  of  conjugate 
hyperbolas.  The  relation- 
ship between  them  is  per- 
fect in  its  duality.  We 
say,  then,  that  each  is  the 
conjugate  of  the  other. 

The  two  hyperbolas  to-  FIG.  13 

gether   are    tangent    exter- 
nally at  their  vertices  to  the  rectangle  of  §  4  at  the  mid-points 
of  its  sides.     Moreover,  all  straight  lines  through  the  common 
center  0,  except  two,  meet  one  hyperbola  or  the  other  in  two 
points,  and  the  segment  thus  terminated  is  bisected  at  0. 


142  ANALYTIC   GEOMETRY 

The  student  should  compare  these  facts  with  the  correspond- 
ing ones  concerning  a  single  ellipse  and  the  circumscribed 
rectangle. 

EXERCISES 

1.  Find  the  coordinates  of  the  foci,  the  lengths  of  the  axes, 
the  slopes  of  the  asymptotes,  and  the  value  of  the  eccentricity 
for  each  of  the  hyperbolas  :' 


(6)  5af-  -  4^  +  20  =  0  ;  (d)  3x^-  -2y*  +  6  =  0. 

Draw  an  accurate  figure  in  each  case. 

2.  What  are  the  equations  of  the  hyperbolas  conjugate  to 
the  hyperbolas  of  Ex.  1  ? 

3.  Find  the  equation  of  the  hyperbola  whose  vertices  are  in 
the  points  (0,  ±  4)  and  whose  eccentricity  is  f  . 

Ans.   4z2-5^  +  80  =  0. 

4.  Find  the  equation  of  the  hyperbola  the  extremities  of 
whose  minor  axis  are  in  the  points  (  ±  3,  0)  and  whose  eccen- 
tricity is  ^. 

5.  Prove  that  the  sum  of  the  squares  of  the  reciprocals  of 
the  eccentricities  of  the  two  conjugate  hyperbolas 

^-UL=l          ^!_l!=  _l 
9      16  9      16 

is  equal  to  unity. 

6.  Prove  the  theorem  of  Ex.  5  for  the  general  pair  of  conju- 
gate hyperbolas. 

7.  Show  that  the  foci  of  a  pair  of   conjugate   hyperbolas 
lie  on  aicircle. 

^V> 

9.  Paip,metric  Representation.  It  is  possible  to  construct  a 
hyperbola,  given  its  axes,  AA'  and  BB',  by  a  method  much 
like  that  of  Ch.  VII,  §  10,  for  the  ellipse. 


THE   HYPERBOLA 


143 


Let  the  two  circles,  C  and  C",  and  the  ray  from  0,  be  drawn 
as  before.  At  the  point  L 
draw  the  tangent  to  C",  and 
mark  the  point  »Q  where  the 
ray  cuts  this  line.  At  R  draw 
the  tangent  to  C  and  mark  the 
point  S  where  this  tangent 
cuts  the  axis  of  x. 

The  locus  of  the  point 
P :  (x,  y),  in  which  the  paral- 
lel to  the  axis  of  x  through 

FIG.  14 
Q    and    the    parallel    to    the 

axis  of  y  through  S  intersect,  is  the  hyperbola. 


For, 
and 

Hence 

and  since 

it  follows  that 


OR  =  a, 

x  =  OS  =  a  sec  <f>, 


OL  =  b, 

y=  LQ  =  b  tan<£. 

*•  =  tan  <f>, 


'sec2  <f>  —  tan2  <£  =  1, 


a 


Conversely,  any  point  (x,  y)  whose  coordinates  satisfy  this 
equation  is  seen  to  lead  to  an  angle  <f>,  for  which  the  above 
formulas  hold. 

We  thus  obtain  the  following  parametric  representation  of 
the  hyperbola : 

x  =  a  sec  </>,        y  =  b  tan  <f>. 

The  circle  C,  constructed  on  the  major  axis  of  the  hyperbola 
as  a  diameter,  is  known  as  the  auxiliary  circle  of  the  hyperbola, 
and  the  angle  </>  is  called  the  eccentric  angle. 

EXERCISES 

1.    Carry  out  the  construction  described  above  for  the  cases  : 
(a)         a  =  3  cm.,         6  =  2  cm. 


144 


ANALYTIC   GEOMETRY 


(6)         a  =  3  cm.,         b  =  3  cm. 
(c)         a  =  2  cm.,         b  =  3  cm. 
2.    Obtain  a  parametric  representation  of  the  hyperbola 

~A*~!&  = 

10.  Conic  Sections.  The  ellipse  (inclusive  of  the  circle),  the 
hyperbola,  and  the  parabola  are  often  called  conic  sections, 

because  they  are  the  curves 
in  which  a  cone  of  revolution 
is  cut  by  planes. 

Suppose  a  plane  M  cuts  only 
one  nappe  of  the  cone,  as  is 
shown  in  the  accompanying 
drawing.  Let  a  small  sphere 
be  placed  in  the  cone  near  O, 
tangent  to  this  nappe  along  a 
circle.  It  will  not  be  large 
enough  to  reach  to  the  plane 
M.  Now  let  the  sphere  grow, 
always  remaining  tangent  to 
the  cone  along  a  circle.  It 
will  finally  just  reach  the 
plane.  Mark  the  point  of 
tangency,  F,  of  the  plane  M 
with  the  sphere,  and  also  the 
FIG-  15  circle  of  contact,  (7,  of  the 

sphere  with  the  cone. 

As  the  sphere  grows  still  larger,  it  cuts  the  plane  M,  but 
finally  passes  beyond  on  the  other  side.  In  its  last  position,  in 
which  it  still  meets  M,  it  will  be  tangent  to  M.  Let  the  point 
of  tangency  be  denoted  by  F',  and  the  circle  of  contact  of  the 
sphere  with  the  cone  by  C'. 

Through  an  arbitrary  point  P  of  the  curve  of  intersection  of 
M  with  the  cone  passes  a  generator  OP  of  the  cone  ;  let  it  cut 
Gin  R  and  C'  in  R.  Then  RR,  being  the  slant  height  of 


THE   HYPERBOLA  145 

the  frustum  *  cut  from  the  cone  by  the  planes  of  C  and  C-,  is 
of  the  same  length,  2  a,  for  all  points  P. 

Join  P  with  F.  Then  PF  and  PR,  being  tangents  from  P 
to  the  same  sphere,  are  equal.  Similarly,  PF1  and  PR'  are 
equal.  Hence 

FP  +  F'P=  RP  +  R'P=  RR', 
or  FP+F'P  =  2a. 

But  this  locus  is  by  definition  an  ellipse  with  its  foci 
at  F  and  F',  and  hence  the  proposition  is  proved  for  the  case 
that  M  cuts  only  one  nappe,  the  intersection  being  a  closed 
curve. 

If  the  plane  M  cuts  both  nappes,  but  does  not  pass  through 
0,  it  is  a  little  harder  to  draw  the  figure,  one  sphere  being 
inscribed  in  the  one  nappe,  the  other,  in  the  other  nappe. 
A  similar  study  shows  that  here  the  difference  between 
FP  and  F'  P  is  equal  to  RR',  and  hence  the  locus  is  a 
hyperbola. 

The  parabola  corresponds  to  the  case  that  M  meets  only  one 
nappe,  but  does  not  cut  it  in  a  closed  curve.  This  case  is 
realized  when  M  does  not  pass  through  0  and  is  parallel  to  a 
generator  of  the  cone. 

Let  L  be  a  line  which  is  perpendicular  to  the  axis  of  the 
cone  in  a  point  of  the  axis  distinct  from  the  vertex.  As  a 
plane,  M,  rotates  about  L,  it  will  cut  from  the  cone  all  three 
kinds  of  conies.  This  will  still  be  true  if  we  take,  as  L,  any 
line  of  space  which  does  not  pass  through  the  vertex  and  is 
not  parallel  to  a  generator. 

11.  Confocal  Conies.  Two  conies  are  said  to  be  confocal  if 
they  have  the  same  foci ;  in  the  case  of  two  parabolas,  we  de- 
mand, further,  that  they  have  the  same  axis. 

*  No  technical  knowledge  of  Solid  Geometry  beyond  the  definitions  of 
the  terms  used  (which  can  be  found  in  any  dictionary)  is  here  needed. 
On  visualizing  the  figure,  the  truth  of  the  statements  regarding  the  space 
relations  becomes  evident. 


146 


ANALYTIC   GEOMETRY 


FIG.  16 


Consider  an   ellipse   and  a  hyperbola  which  are   confocal. 
They  evidently  intersect  in  four  points.* 

Let  P  be  one  of  these  points.  Join  P  with  F  and  F'. 
Then  FP  and  F'P  are  focal  radii  both  of  the  ellipse  and  of  the 
hyperbola.  Now,  the  tangent  to  a  hyper- 
bola at  any  point  not  a  vertex  bisects  the 
angle  between  the  focal  radii  drawn  to 
that  point,  §  5 ;  and  the  normal  to  an 
ellipse  at  any  point  not  on  the  transverse 
axis  bisects  the  angle  between  the  focal 
radii  drawn  to  that  point,  Ch.  VII,  §  4. 
It  follows,  then,  that  the  tangent  to 
the  hyperbola  at  P  and  the  normal  to  the  ellipse  at  this 
point  coincide.  Hence  the  two  curves  intersect  at  right 
angles,  or  orthogonally,  as  we  say.  We  have  thus  proved  the 
following 

-  THEOREM.     A  pair  of  confocal  conies,  one  of  which  is  an  el- 
lipse and  the  other  a  hyperbola,  cut  each  other  orthogonally. 

Confocal  Parabolas.  Consider  two  parabolas  having  the 
same  focus  and  the  same  axis.  If  both  open  out  in  the  same 
direction,  they  have  no  point  in  common.  If,  however,  they 
open  out  in  opposite  directions,  they  intersect  in 
two  points  which  are  symmetrically  situated  with 
respect  to  the  axis. 

In  the  latter  case,  the  parabolas  intersect  orthogo- 
nally, as  has  already  been  proved  analytically ;  cf . 
Ch.  VI,  §  3,  Ex.  10. 

This  result  could  have  been  forecast,  as  a  conse- 
quence of  the  relations  established  in  §  7.     For,  if 
one  focus,  F,  and  the  two  corresponding  directrices  of  a  pair 
of  confocal  conies,  consisting  of  an  ellipse  and  a  hyperbola, 
are  held  fast,  and  if  the  other  focus  is  made  to  recede  in- 
definitely, each  of  the  conies  approaches  a  parabola.     But  the 


FIG.  17 


*  Let  the  student  satisfy  himself  that  two  confocal  ellipses  do  not  in- 
tersect, and  that  the  same  is  true  of  two  confocal  hyperbolas. 


THE    HYPERBOLA 


147 


FIG.  18 


conies  always  intersect  orthogonally,  and  so  the  same  will  be 
true  of  the  limiting  curves,  the  parabolas. 

To  obtain  a  prescribed  pair  of  parabolas,  like  those  described 
above,  as  limiting  curves,  it  is  necessary  merely  to  choose  the 
two  confocal  conies  so  that  the  directrices  corresponding  to  F 
are  at  the  proper  distances  from  F. 

Mechanical  Constructions.     It  is  possible  to  draw  with  ease  a 

large  number  of  confocal  ellipses  by  the  method  set  forth  in  Ch. 

VII,  §  1.     Let  thumb  tacks  be  inserted  at  F 

and  F',  but  not  pushed  clear  down.     Let  a 

thread  be  tied  to  the  tack  at  F,  passed  round 

the  tack  at  F',  and  held  fast  at  M.    Then  an 

ellipse  can  be  drawn  with  F  and  F'  as  foci. 
Now  let  the  thread  be  unwound  at  F' 

and  drawn  in  or  paid  out  slightly,  so  that 

the  length  of  the  free  thread  between  F  and  F'  is  changed. 

On  repeating  the  above  construction,  a  second  ellipse  with 

its  foci  at  F  and  F'  is  obtained  ;  and  so  on. 

There  is  an  analogous  construction  for  a  hyperbola,  which 

has  not  yet  -been  mentioned.  Tie  a  thread  to  a  pencil  point,* 
pass  the  thread  round  the  pegs  at  F  and 
F'  as  shown,  hold  the  free  ends  firmly 
together  at  M,  and,  keeping  the  thread 
taut  by  pressing  on  the  pencil,  allow  M  to 
move.  The  pencil  then  obviously  traces 
out  a  hyperbola. 

By  pulling  one  end  of  the 
thread  in  slightly  at  M,  or  by 

paying  it  out,  and  then  repeating  the  construction, 

a  new  hyperbola  with  the  same  foci  is  obtained  ; 

and  so  on. 

Parabolas.     The  accompanying  figure  suggests 

a  means  for  drawing  a  parabola  mechanically. 

*  To  keep  the  thread  from  slipping  off,  cut  a  groove  in  the  lead,  such 

as  would  be  obtained  if  the  pencil  were  turned  about  its  axis  in  a  lathe 

and  the  point  of  a  chisel  were  held  against  the  lead  close  to  the  wood. 


FIG.  19 


FIG.  20 


148  ANALYTIC  GEOMETRY 

A  ruler,  D,  is  held  fast  and  a  triangle,  T,  is  allowed  to  slide 
along  the  ruler.  A  thread  is  tied  at  F  and  Q,  and  a  pencil 
point,  P,  keeps  the  thread  taut  and  pressed  against  the 
triangle. 

EXERCISES 

1.  Show  that  the  conies, 

*+*_!    and     |_g=1, 

are  confocal. 

2.  Prove  that  the  equation, 


9+A     5+A 

represents  an  ellipse  for  each  value  of  A  greater  than  —  5  and 
represents  a  hyperbola  for  each  value  of  A  between  —  9  and 
—  5.  Show  that  all  these  ellipses  and  hyperbolas  are  confocal, 
with  the  points  (±  2,  0)  as  foci. 

3.  For  what  values  of  A  does  the  equation 

a2  +  x  +  &2  +  A=    ' 

where  a  and  b  are  given  positive  constants  such  that  a  >  6, 
represent  i)  ellipses  ?  ii)  hyperbolas  ?  Show  that  all  these 
conies  are  confocal. 

4.  Draw  a  set  of  confocal  ellipses  and  hyperbolas. 

5.  Draw  a  set  of  confocal  parabolas,  all  having  the  same 
transverse  axis,  some  opening  in  one  direction,  some  in  the 
other. 

EXERCISES  ON  CHAPTER  VIM 

1.  The  axes  of  a  hyperbola  which  goes  through  the  points 
(1,  4),  (—  2,  7)  are  the  axes  of  coordinates.     Find  the  equation 
of  the  hyperbola.  Ans.   y2  —  llo;2  =  5. 

2.  Show  that  the  hyperbolas  denned  by  the  equation 

4#2  —  5  y1  =  c, 


THE    HYPERBOLA  149 

where  c  is  an  arbitrary  constant,  not  zero,  all  have  the  same 
asymptotes. 

3.   How  many  hyperbolas  are  there  with  the  lines 
3  a?— 


as  asymptotes  ?     Find  an  equation  which  represents  them  all. 

Ans.   3  x1  —  16  y1  =  c,     c  =£  0. 

4.  What  is  the  equation  of  all  the  rectangular  hyperbolas 
with  the  axes  of  coordinates  as  axes? 

5.  A  hyperbola  with  the  lines  4  x1  —  y"1  =  0  as  asymptotes 
goes  through  the  point  (1,  1).     What  is  its  equation  ? 

Ans.   4  x~-  —  y2  =  3. 

6.  The  asymptotes  of  a  hyperbola  go  through  the  origin 
and  have  slopes  ±  2.     The  hyperbola  goes  through  the  point 
(1,  3).     Find  its  equation.  Ans.   4^-  —  y2  =  —  5. 

7.  The   two  hyperbolas  of  Exs.  5  and  6  have  the  same 
asymptotes,  but  lie  in  the  opposite  pairs  of  regions  into  which 
the  plane  is  divided  by  the  asymptotes.     Show  that  the  sum 
of  the  squares  of  the  reciprocals  of  their  eccentricities  equals 
unity. 

8.  Prove  that  of  the  hyperbolas  of  Ex.  2  those  for  which 
c  is  positive  are  all  similar,  and  that  this  is  true  also  of  those 
for  which  c  is  negative.     If  e  is  the  common  value  of  the  ec- 
centricity of  the  hyperbolas  of  the  first  set  and  e'  is  that  of 
the  hyperbolas  of  the  second  set,  show  that 

(i)  I+l-i. 

e2     e'2 

9.  Prove  that  the  relation  (1)  is  valid  for  the  eccentricities 
of  any  two  hyperbolas  which  have  the  same  asymptotes  but 
lie  in  the  opposite  regions  between  the  asymptotes. 

10.  Show  that  two  hyperbolas  which  are  related  as  those 
described  in  the  previous  exercise  have  the  same  eccentricity 
if  and  only  if  they  are  rectangular  hyperbolas. 


150  ANALYTIC   GEOMETRY 

11.  A  hyperbola  with  its  center  in  the  origin  has  the  eccen. 
tricity  2.     Find  the  equations  of  the  asymptotes,  (a)  if  the 
foci  lie  on  the  axis  of  x  ;  (6)  if  the  foci  lie  on  the  axis  of  y, 

Ans.    (a)  3a?-y'  =  0;  (6)  x"--3y'i  =  0. 

12.  What  is  the  equation   representing  all  the  hyperbolas 
which   have  their  centers  in  the  origin  and   eccentricity  2, 
(a)  if  the  foci  lie  on  the  axis  of  x  ?    (&)  if  the  foci  lie  on  the 
axis  of  y?     Show  that  in  either  case  the  vertices  lie  midway 
between  the  center  and  the  foci. 

13.  Prove  that  the  vertices  of  the  hyperbola 


subtend  a  right  angle  at  each  of  the  points  (0,  ±  6)  when  and 
only  when  the  hyperbola  is  rectangular.  What  is  the  corre- 
sponding theorem  in  the  case  of  the  ellipse  ? 

14.  The  projections  of  a  point  P  of   a   hyperbola  on  the 
transverse  and  conjugate  axes  are  PI  and  P2.     The  tangent  at 
P  meets  these  axes  in  2\  and  T2.     Show  that  OP±  •  07\=  o2 
and  OP2  •  OT2=—bfi,  where  0  is  the  center  of  the  hyperbola 
and  a  and  b  are  the  semi-axes. 

15.  Prove  that  the  segment  of  a  tangent  to  -a  hyperbola  be- 
tween the  point  of  contact  and  a  directrix  subtends  a  right 
angle  at  the  corresponding  focus.    . 

16.  The   projection  of  a   point  P  of  a  hyperbola   on   the 
transverse  axis  is  Pt  and  the  normal  at  P  meets  this  axis  at 
NI.     Show  that  the  ratio  of  the  distances  of  the  center  from 
N!  and  PI  equals  the  square  of  the  eccentricity. 

17.  Prove  that  the  line  joining  a  point  P  of  a  hyperbola 
with  the  center  and  the  line  through  a  focus  perpendicular  to 
the  tangent  at  P  meet  on  a  directrix. 

18.  Find  the  equation  of  the  circle  which  is  tangent  to  a 
hyperbola  at  the  upper  ends  of  the  two  latera  recta. 

19.  Let  0  be  the  center,  A  a  vertex,  and  F  the  adjacent 
focus  of  a  hyperbola.     The  tangent  at  a  point  P  meets  the 


THE   HYPERBOLA  151 

transverse  axis  at  T  and  the  tangent  at  A  meets  OP  at  V. 
Show  that  TV  is  parallel  to  AP. 

20.  Show  that  an  asymptote,  a  directrix,  and  the  line  through 
the  corresponding   focus   perpendicular  to  the  asymptote   go 
through  a  point. 

21.  A  line  through  a  focus  F  parallel  to  an  asymptote  meets 
the  hyperbola  at  P.     Show  that  the  tangent  at  P,  the  other 
asymptote,  and  the  line  of  the  latus  rectum  through  F  meet  in 
a  point. 

22.  Let  F  be  a  focus  and  D  the  corresponding  directrix  of 
a  hyperbola.     A  line  through  a  point  P  of  the  hyperbola  parallel 
to  an  asymptote  meets  D  in  the  point  K.     Prove  that  the  tri- 
angle FPK  is  isosceles. 

Exercises  23-33.  In  proving  the  theorems  in  these  exercises 
it  will,  in  general,  be  necessary  to  make  actual  use  of  the 
equation  expressing  the  fact  that  a  certain  point  lies  on  the 
hyperbola. 

23.  The  tangent  to  a  hyperbola  at  a  point  P  meets  the  tan- 
gent at  one  vertex  in  Q.     Prove  that  the  line  joining  the  other 
vertex  to  P  is  parallel  to  the  line  joining  the  center  to  Q. 

|24?)  Let  F  be  a  focus  and  D  the  corresponding  directrix  of  a 
hyperbola.  Prove  that  the  segment  cut  from  D  by  the  lines 
joining  the  vertices  with  an.  arbitrary  point  on  the  hyperbola 
subtends  a  right  angle  at  F. 

25.  Prove  that  the  product  of  the  distances  of  the  foci  of  a 
hyperbola  from  a  tangent  is  constant,  i.e.  independent  of  the 
choice  of  the  tangent. 

26.  Let  A  and  A'  be  the  vertices  of  a  rectangular  hyperbola 
and  let  P  and  P'  be  two  points  of  the  hyperbola  symmetric  in 
the  transverse  axis.     Prove  that  AP  is  perpendicular  to  A'P1 
and  that  AP  is  perpendicular  to  A'P. 

27.  Show  that  the  product  of  the  focal  radii  to  a  point  on  a 
rectangular  hyperbola  is  equal  to  the  square  of  the  distance  of 
the  point  from  the  center. 


152  ANALYTIC   GEOMETRY 

28.  Prove  that  the  angles  subtended  at  the  vertices  of  a 
rectangular  hyperbola  by  a  chord  parallel  to  the  conjugate  axis 
are  supplementary. 

29.  Prove  that  the  product  of  the  distances  of  an  arbitrary 
point  on  a  hyperbola   from  the  asymptotes   is   constant,  i.e. 
the  same  for  every  choice  of  the  point. 

30.  A  line   through   an  arbitrary  point  P  on  a  hyperbola 
parallel  to  the  conjugate  axis  meets  the  asymptotes  in  M  and 
-2V.     Show  that  the  product  of  the  segments  in  which  P  divides 
MN  is  constant. 

31.  Prove  that  the  segment  of  a  tangent  to  a  hyperbola  cut 
out  by  the  asymptotes  is  bisected  by  the  point  of  contact  of 
the  tangent. 

32.  Show  that  the  tangent  to  a  hyperbola  at  an  arbitrary  point 
forms  with  the  asymptotes  a  triangle  which  has  a  constant  area. 

33.  The  tangent  to  a  hyperbola  at  a  point  P  meets  the  tan- 
gents at  the  vertices  in  M  and  N.     Prove  that  the  circle  on 
MN  as  a  diameter  .passes  through  the  foci. 

Loci 

34.  Find  the  locus  of  a  point  whose  distance  from  a  given 
circle  always  equals  its  distance  from  a  given  point  without 
the  circle.     First  give  a  geometric  construction,  with  ruler  and 
compass,  for  points  on  the  locus.     Then  find  the  equation  of 
the  locus. 

35.  The  base  of  a  triangle  is  fixed  and  the  product  of  the 
tangents  of  the  base  angles  is  a  negative  constant.,    What  is 
the  locus  of  the  vertex? 

36.  A  line  moves  so  that  the  area  of  the  triangle  which  it 
forms  with  two  given  perpendicular  lines  is  constant.     Find 
the  locus  of  the  mid-point  of  the  segment  cut  from  it  by  these 
lines. 

Ans.   Two  conjugate  rectangular  hyperbolas,  with  the  given 
lines  as  asymptotes. 


THE   HYPERBOLA  153 

37.  Given  a  fixed  line  L  and  a  fixed  point  A,  not  on  L.     A 
point  P  moves  so  that  its  distance  from  L  always  equals  the 
distance  AQ,  where  Q  is  the  foot  of  the  perpendicular  dropped 
from  P  on  L.     What  is  the  locus  of  P? 

38.  What  is  the  locus  of  the  point  P  of  the  preceding  exer- 
cise, if  the  ratio  of  its  distance  from  L  to  the  distance  AQ  is 
constant  ? 


CHAPTER   IX 
CERTAIN   GENERAL   METHODS 

1.  Tangents.  Let  it  be  required  to  find  the  tangent  line  to 
a  given  curve  at  an  arbitrary  point. 

In  the  case  of  the  circle  the  tangent  is  perpendicular  to  the 
radius  drawn  to  the  point  of  tangency.  But%this  solution  is 
of  so  special  a  nature  that  it  suggests  no  general  method  of 
attack.  A  general  method  must  be 
based  on  a  general  property  of  tan- 
gents, irrespective  of  the  special  curve 
considered.  Such  a  method  is  the 
following.  Let  P  be  an  arbitrary 
point  of  a  given  curve,  (7,  at  which  it 
is  desired  to  draw  the  tangent,  T.  Let 
a  second  point,  Pf,  be  chosen  on  (7, 
and  draw  the  secant,  PP'.  As  P' 
moves  along  C  and  approaches  the 

fixed  point  P  as  its  limit,  the  secant  rotates  about  P  as  a  pivot 
and  approaches  the  tangent,  T,  as  its  limiting  position.  Thus 
the  tangent  appears  as  the  limit  of  the  secant. 

If,  now,  in  a  given  case  we  can  find  an  expression  for  the 
slope  of  the  secant,  the  limit  approached  by  this  expression 
will  give  us  the  slope  of  the  tangent.  The  slope  of  the  tangent 
to  the  curve  at  P  we  shall  call,  for  the  sake  of  brevity,  the 
slope  of  the  curve  at  P. 

Example  1.     Find  the  slope  of  the  curve 
(1)  y  =  tf 

at  a  given  point,  P. 

154 


FIG.  1 


CERTAIN  GENERAL  METHODS 


155 


Let  the  .coordinates  of  P  be 
(zi,  2/1)  ;  those  of  P',  («',  y'), 
or  (a?!  -f  /t,  2/1  +  ft).  Then 


and  we  have,  for  the  slope  of 
the  secant  PP\  the  expression  : 


(2) 


i 
tanr  =  -, 


FIG.  2 


where  T'  =  £  QPP'.     The  slope  of  the  tangent  line,  T7,  at  P  is, 
then, 

A; 
(3)  ' 


tan  T  =  lira  tan  r'  =  lim  -  , 

P'=P  h±0  h 


where  T  =  *$.  QPT.     The  sign  =  is  used  to  mean  "approaches 

as  its  limit,"  and  the  expression  :    lim  tan  T',  is  read  :    "  the 

P'=P 

limit  of  tan  T',  as  P'  approaches  P." 

Suppose,  for  example,  that  P  is  the  point  (1,  1).  Let  us 
compute  A;  and  tan  T'  for  a  few  values  of  h.  Here,  Xi  =  1  and 
?/!  =  1.  If  h  =  .1,  then 

z'  =  a?!  +  £  =  1.1, 


and  hence 


k  =  .21, 
=  ^  =  2.1. 


Next,  let  P'  be  the  point  for  which 

x'  =1.01. 
Then  y'  =  1.0201, 

h  =  .01,  ft  =  .0201, 

.0201 


and  hence 


tan  T'  = 


.01 


=  2.01. 


Let  the  student  work  out  one  more  case,  taking  x'  =  1.001. 
He  will  find  that  here  k  =  .002001  and 


tan  T'  =  2.001. 


156 


ANALYTIC   GEOMETRY 


These  results  can  be  presented  conveniently  in  the  form  of 
a  table : 


k 

h 

k 

tanr'  =  - 

h 

.1 

.21 

2.1 

.01 

.0201 

2.01 

.001 

.002001 

2.001 

The  numbers  in  the  last  column  appear  to  be  approaching 
nearer  and  nearer  to  the  limit  2  ;  in  other  words,  the  slope  of 
the  curve  in  the  point  (1,  1)  appears  to  be  2.  Let  us  prove 
that  this  is  actually  the  case.  Since  the  proof  is  just  as  simple 
for  an  arbitrary  point  P,  we  will  return  to  the  general  case. 

The  point  P  being  a  point  of  the  curve  (1),  its  coordinates 
(a?!,  i/i)  must  satisfy  that  equation.  Hence 

(4)  yi  =  xf. 

Similarly,  for  the  point  P'  whose  coordinates  are  (xi+h, 


or 

(5)  yl+k  =  xl2  +  2  xji 

Subtracting  (4)  from  (5),  we  get  : 


Consequently, 


Now  let  P'  approach  P;  h  will  then  approach  0,  and  we 
shall  have 


k 
lim  tan  T'  =  lim  -  =  lim  (2  xl  +  K). 

P'=P  Ad=0    h  h^O 


But 


lim  tan  T'  =  tan  T,         and 
Heace  tan  T  =  2  xv 


CERTAIN  GENERAL  METHODS        157 

We  can  say,  then,  that  the  slope  of  the  curve  (1),  at  an  arbi- 
trary point  P :  (xi,  y^)  on  it,  is 

X  =  2x{. 

If,  in  particular,  P  is  the  point  (1,  1),  the  slope  of  the  tan- 
gent there  is  A.  =  2  •  1  =  2,  and  thus  the  indication  given  by 
the  above  table  is  seen  to  be  borne  out. 

Example  2.     Find  the  slope  of  the  curve 
(6)  ,-£ 

at  an  arbitrary  point  P :.  (xt,  y^)  of  the  curve. 

Denote,  as  before,  the  coordinates  of  a  second  point,  P',«by 

x'  =  ajj.  +  h,  yf  =  yl  +  k. 

Then,  since  P  and  P'  lie  on  the  curve, 


and 
Hence 


h 
a2 


Nothing  is  more  natural  than  to  reduce  the  right-hand  side 
of  this  equation  to  a  common  denominator.     Thus 

^  __         (i  fa    t 
Xi(xi  -j-  K) 
Consequently, 

,      k          —a2 
tan  T  =-  =  • 
h 


We  are  now  ready  to  let  P'  approach  P: 

—  a2 


lim  tan  T'  =  lim 

P'±P  »=o  x^Xi  +  h) 

The   limit  approached   by  the   right-hand   side  is  obviously 
—  a2/#!2,  and  so 

a2 

tan  T  =  --- 


158  ANALYTIC   GEOMETRY 

We  have,  then,  as  the  final  result :   The  slope  of  the  curve  (6), 
at  an  arbitary  point  (x1}  y^)  on  it,  is 

\  =  -^. 

Equation  of  the  Tangent.     Since  the  tangent  to  the  curve  (1), 

at  the  point  (1,  1)  has  the  slope  2,  its  equation  is 

y-l=2(x-l),          or         2x-y-l=0. 

Similarly,  the  equation  of  the  tangent  to  the  curve  (1)  at  an 
arlfitrary  point  P  :  (x1}  y^  is 

or 


This  equation  may  be  simplified  by  use  of  the  equality, 

y\  =  «i2, 

which  says  that  the  point  P  lies  on  the  curve.  For,  if  we  re- 
place the  term  2  x^  by  its  equal,  2  y1}  and  then  combine  the 
terms  in  y1}  the  equation  becomes 


This  equation  of  the  tangent  is  of  the  first  degree  in  x  and  y, 
as  it  should  be.  The  quantities  a^  and  y±  are  the  arbitrary, 
but  in  any  given  case  fixed,  coordinates  of  P  and  are  not 
variables. 

Equation  of  the  Normal.  The  line  through  a  point  P  of  a 
curve  perpendicular  to  the  tangent  at  P  is  known  as  the 
normal  to  the  curve  at  P. 

Since  the  tangent  to  the  curve  y  =  x2  at  the  point  (1,  1) 
has  the  slope  2,  the  normal  at  this  point  has  the  slope  —  -|. 
Consequently,  the  equation  of  the  normal  is 


or 


CERTAIN  GENERAL  METHODS  159 

EXERCISES 

1.  Determine  the  slope  of  the  curve  y  =  x2  —  x  at  the  point 
(3,  6).     First  make  out  a  table  like  that  under  Example  1,  and 
hence  infer  the  probable  slope.     Then  take  an  arbitrary  point 
(xi>  y\)  on  the  curve  and  determine  the  actual   slope  at  this 
point  by  finding 

v     k 
lim  — • 

*=y)  h 

2.  The  same  for  the  curve  8y  =  3x*  at  the  point  (2,  3). 
\$.    The  same  for  the  curve  y  =  2xz  —  3  a;  + 1  at  the  point 

(1,  0). 

Find  the  slope  of  each  of  the  following  curves  at  an  arbitrary 
point  P :  (xi,  y^).  No  preliminary  study  of  a  numerical  case, 
like  that  which  gave  rise  to  the  table  under  Example  1,  is 
here  required. 

4.   y  =  x"  —  3 x  +  1.  Ans.   X  =  2xl  —  3- 


6.    y  =  y?  —  x.  8.    y  =  x3  +  x"-  +  cc  +  1. 

=  ic'+px-l-  q.  Ans.   \  =  3x12 

10.   y  =  x*  —  a*.  Ans.    A  =  4or13. 


i3.   Jf——  —  Ans.   \  = 


l-x  (1- 

Q  rf  Q 

t   A                                    &  •"  A                    \                                        O 

14-    y  =  ^ :•  Ans.    A  =  —  — 


/  =  ace"  +  60;  4-  c.  u4«s.   X  =  2  a^!  +  &. 

16.   y  =  ax3  +  6x7  -\-cx-\-d. 

.    ?/  =  a;",  (n,  a  positive  integer)  u4?^s.    X  =  no?!""1. 

18.      =  cxn. 


160  ANALYTIC   GEOMETRY 

Find  the  equations  of  the  tangents'  to  the  following  curves 
at  the  points  specified.  In  each  case  reduce  the  equation  ob- 
tained to  the  simplest  form. 

19.  The  curve  of  Ex.  1  at  the  points  (3,  6)  ;     (ajl5  y^). 

Ans.   5x  —  y  —  9  =  0;         (2  a?t  —  T)x  —  y  —  x?  =  0. 

20.  The  curve  of  Ex.  3  at  the  points  (xl9  y^)  ;     (1,  0). 

21.  The  curve  of  Ex.  4  at  the  points  (a^,  y^)  ;     (—  1,  5). 

22.  The  curve  of  Ex.  11  at  the  points  (1,  1)  ;     (xl}  3^). 

23.  The  curve  of  Ex.  17  at  the  point  (xl}  y^. 

Ans.   nxin~lx  —  y—(n  —  l)yl  =  0. 

24.  The  curve  of  Ex.  13  at  the  point  whose  abscissa  is  2. 

25.  The  curve  of  Ex.  14  at  the  point  whose  abscissa  is  4. 

26.  Find  the  equations  of  the  normals  to  the  curves  of  Exs. 
21,  22  at  the  designated  points, 

2.  Continuation.  Implicit  Equations.  We  have  applied  the 
general  method  to  curves  whose  equations  are  given  in  the 
form  :  y  =  a  simple  expression  in  x.  More  precisely,  this 
"  simple  expression  "  has  each  time  been  a  polynomial  (or  even 
a  monomial),  or  the  ratio  of  two  such  expressions. 

But  even  the  simplest  forms  of  the  equations  of  the  conies 
are,  as  a  rule,  such  that,  if  the  equation  be  solved  for  y,  radi- 
cals will  appear.  In  such  cases,  the  following  method  of  treat- 
ment can  be  used  with  advantage. 

The  Parabola.     Let  it  be  required  to  find  the  slope  of  the 
parabola 
(1)  y* 


at  any  point  P  :  (xl}  y^  on  the  curve. 

We  will  treat  first  a  numerical  case,  setting  m  =  2: 

(2)  jf  =  4o. 
Since  P  is  on  the  curve,  we  have 

(3)  2^  =  4^. 


CERTAIN  GENERAL  METHODS        161 

Since  P'  :  (xt  +  h)yl-\-  k)  is  also  on  the  curve,  we  have  : 


or 

(4)  yi2  +  2  yjc  +  V  =  4*!  -f  4  A. 

Subtract  (3)  from  (4)  : 


Divide  this  equation  through  by  h,  to  obtain  an  equation  for 
tan  T'  =  k/h : 

k  ,   ,  k 

2  2/1  -  4-  k  -  =  4,         or        2 
h         h 

Solve  the  latter  equation  for  tan  T' 
tan  /  =  — 


We  are  now  ready  to  let  P'  approach  P  as  its  limit.  This 
means  that  h  and  k  both  approach  0.  We  have,  then, 

lim  tan  T'  ;=  lirn , 

P'±P  A=O  2  y-i  +  k 

or 

4        2 

tan  T  = =  —  • 

2  2/i     2/i 

It  has  been  tacitly  assumed  that  y^  3=  0.  If  yt  =  0,  then 
tanr'  increases  indefinitely  as  h,  and  with  it  k,  approaches 
zero.  Thus  the  tangent  line  is  seen  to  be  perpendicular  to  the 
axis  of  x  at  this  point,  as  obviously  is,  in  fact,  the  case,  since 
the  point  is  the  vertex  of  the  parabola. 

The  student  will  now  carry  through  by  himself  the  corre- 
sponding solution  in  the  general  case  of  equation  (1).  He 
will  arrive  at  the  result :  The  slope  A  of  the  parabola 

y1*  =  2  mx 
at  an  arbitrary  point  (xv,  y^  of  the  curve  is 

(5) 


162  ANALYTIC   GEOMETRY 

The  Ellipse.     The  treatment  in  the  case  of  the  ellipse, 


is   precisely   similar.     Writing    (6),  for   convenience,   in   the 
form 

(7)  &"»"  -f-  a'y"  =  a"bn, 

we  are  led  to  the  following  equations  *  : 

(8)  Wx?  +  aV  =  a2&2  ; 


or 

(9)        b^xj2  +  a2!/!2  +  2  IPxJi  +  2a?yik  +  62/i2  +a2fc2  =  a262. 

Subtract  (8)  from  (9)  : 

2  b^h  +  2  tfyjc  +  bW  +  a%2  =  0. 
Divide  by  h  : 

-  =  0, 


h  h 

or  2  b^  +  2  a22h  tan  T'  +  62/i  +  a2A;  tan  r'  =  0. 

Solve  this  equation-  for  tan  T'  : 

tan  T'  = 


+  a2fc 
Now  let  P'  approach  P  as  its  limit  : 

,      ,. 
hm  tan  r'  =  lim  — 


Hence  tan  T  =  — 

We  have  thus  obtained  the  result  :   The  slope  X  of  the  ellipse 
x2     y!=1 
a2     62 
at  an  arbitrary  one  of  its  points  (xt,  y^)  is 

(10)  X  =  -^i- 

0^1 

*  The  student  will  do  well  to  paraphrase  the  text  at  this  point  with  a 
numerical  case,  —  say,  4x2  +  9j/2  =  36. 


CERTAIN  GENERAL  METHODS        163 

The  Hyperbola.     The  treatment  is  left  to  the  student.     The 
result  is  as  follows. 

The  slope  \  of  the  hyperbola 

?_2_2/?=l 
a2     62 

at  an  arbitrary  one  of  its  points  (x1}  y^)  is 

(11)  X  =  ^. 

<fyi 

Equation  of  the  Tangent.     Since  the  slope  of  the  ellipse  at 
the  point  (x^  yi)  is  —  fe^/a2^,  the  equation  of  the  tangent  at 
*          is 


or,  after  clearing  of  fractions  and  rearranging  terms, 


If  we  divide  both  sides  of  this  equation  by  a262,  we  have 

ftifl  -  .Vi.V__^i2  .  y\z 
a?  ^  52      a2      62  ' 

But,  since  the  point  (xly  y^)  lies  on  the  ellipse,  it  follows  that 


, 
a2       ft2 

and  the  equation  of  the  tangent  becomes 
xix  i  .Viy  - 


equation  of  the  tangent  to  the  ellipse 

£+<-l 

a2     62 


a<  ^e  pom?  (xj, 
(12) 


a2       62 


In  a  similar  manner  let  the  student  establish  the  equations 
of  the  tangents  to  the  hyperbola  and  the  parabola. 


164  ANALYTIC   GEOMETRY 


a2      62 
at  the  point  (x1?  y^)  has  the  equation 

rt%  7i2 

tangent  to  the  parabola 

y*  =  2  maj 
Ci,  f/i)  ^««  ^e  equation 

/•i  A\  ..  -.       j  rm(w  -I-  /*•  \ 

EXERCISES 

Find  the  slope  of  each  of  the  following  six  curves  at  an 
arbitrary  one  of  its  points,  applying  each  time  the  method 
set  forth  in  the  text. 

1.  2#2  +  3?/2  =  12.  3.   yz  =  12x. 

2.  x2  —  4y2  =  4.  4.   cc2  —  t/2  =  a2. 

5.  ^4x2  +  .Bi/2  =  C,  where  A,  B,  C  are  all  positive. 

6.  ?/2  =  Ax  +  S,  where  A  4=-  0. 

7.  Find  the  slope  of  the  parabola  yz  +  2y  =  6 a;  at  the  point 

V       1)     Vl/"  •^J.rvS*        ^   *—    • 

^1  +  1 

8.  What  is  the  slope  of  the  parabola  of  Ex.  7  at  the  origin? 

Ans.   3. 

9.  Find  the  slope  of  the  curve 


at  the  origin.  Ans.   X=  f . 

Suggestion.     First  find  the  slope  at  an  arbitrary  point  (xy,  y^). 
Then  substitute  in  the  result  the  coordinates  of  the  origin. 

10.   What  angle  does  the  curve 

make  with  a  parallel  to  the  axis  of  x  at  the  point  (1, 1)  ? 

Ans.  45°. 


CERTAIN  GENERAL  METHODS 


165 


11.  Find  the  slope  of  the  curve  xy  =  a2  at  any  point  (xlf  yt) 
by  the  method  of  the  present  paragraph,  and  show  that  your 
result  agrees  with  that  of  §  1,  Example  2. 

Find  the  equation  of  the  tangent  to  each  of  the  following 
curves  at  the  point  designated,  applying  each  time  the  method 
of  the  text.  Reduce  the  equation  to  its  simplest  form. 

12.  The  curve  of  Ex.  1  at  the  point  (xi}  yi). 

Ans.   2  X]X  +  3  y^y  =  12. 

13.  The  curve  of  Ex.  3  at  the  points  (xl}  y^) ;     (3,  —  6). 

14.  The  curve  of  Ex.  5  at  the  point  (xt,  y^. 

Ans.   Ax-p  +  By$  =  C. 

15.  The  curve  of  Ex.  6  at  the  point  (a?i,  yi). 

16.  The  curve  of  Ex.  7  at  the  points  (xi}  y^  ;     (£,  1). 

17.  The  curve  of  Ex.  9  at  the  origin. 

18.  Find  the  equations  of  the  normals  to  the  curves  of  Exs. 
12,  13  at  the  points  specified. 


\ 


3.  The  Equation  u  +  kv  =  0.  Consider  the  following  ex- 
ample. 

The  equations 

(1)  x  +  y-2  =  0, 

(2)  x-y  =  0, 

represent  two  straight  lines  intersecting 
in  the  point  (1,  1),  as  shown  in  Fig.  3. 
What  can  we  say  concerning  the  curve  * 

(3)  (x  +  y-2)+k(x-y)=0, 

where  k  denotes  a  constant  number  ? 

This  curve  is  a  straight  line,  since  (3)  is  an  equation  of  the 
first  degree  in  x  and  y.  Suppose,  now,  that  various  different 
values  are  given  to  k.  Then  (3)  represents  various  straight 
lines  in  turn.  What  do  all  these  lines  have  in  common  ? 

*  The  word  "curve"  is  used  here  in  the  sense  common  in  analytic 
geometry,  to  denote  merely  the  "  locus  of  the  equation."  Consequently 
a  curve  in  this  sense  is  not  necessarily  crooked  ;  it  may  be  a  straight  line. 


FIG.  3 


166  ANALYTIC   GEOMETRY 

Since  the  lines  (1)  and  (2)  intersect  in  the  point  (1,  1),  the 
coordinates  of  this  point  make  the  left-hand  sides  of  equations 
(1)  and  (2),  namely,  the  expressions, 

x  +  y  —  2         and        x  —  y, 

vanish.  Consequently,  they  always  make  the  left-hand  side 
of  equation  (3)  vanish.  In  other  words,  equation  (3)  is  satisfied 
by  the  coordinates  of  the  point  of  intersection  of  the  lines  (1)  and 
(2),  NO  MATTER  WHAT  VALUE  Jc  HAS.  This  means  that  all  the 
straight  lines  represented  by  (3)  go  through  the  point  of  inter- 
section of  the  lines  (1)  and  (2). 

The  result  can  be  restated  in  the  following  form.  Let  the 
single  letter  u  stand  for  the  whole  expression  x  +  y  —  2 : 

u  =  x  +  y  —  2, 

the  sign  =  meaning  identically  equal,  i.e.  equal,  no  matter  what 
values  x  and  y  have.  Similarly,  let  v  stand  for  x  —  y : 

v  =  x  —y. 

Then  (3)  takes  on  the  form : 
(4)  u  +  kv  =  0. 

We  now  restate  our  result. 

Ifu  =  0  and  v  =  0  are  the  equations  of  two  intersecting  straight 
lines,  then  the  equation 

u  +  kv  =  0 

represents  a  straight  line  which  goes  through  the  point  of  inter- 
section of  the  two  given  lines. 

By  giving  to  k  a  suitable  value,  u  +  kv  =  0  can  be  made  to 
represent  any  desired  line  through  the  point  of  intersection 
(XD  y\)  °f  the  given  lines,  with  the  sole  exception  of  the  line 
v  =  0.  For,  let  L  be  the  desired  line,  and  let  (ccj,  t/2)  be  a 
point  of  L  distinct  from  (x^  y^.  Then,  on  substituting  for  x 
and  y  the  values  x2  and  y2  in  the  equation  u  -J-  Jcv  =  0,  we  ob- 
tain an  equation,  in  which  Jc  is  the  unknown.  This  equation 
can  be  solved  for  k,  since  v  does  not  vanish  for  the  point 


CERTAIN  GENERAL  METHODS        167 

Example.  Find  the  equation  of  the  line  L  which  goes 
through  the  point  of  intersection  of  the  lines  (1)  and  (2)  and 
cuts  the  axis  of  y  in  the  point  (0,  —  4). 

The  required  line,  L,  is  one  of  the  lines  (3) ;  i.e.  for  a  suit- 
able value  of  k,  (3)  will  represent  L.  To  find  this  value  of  A:, 
we  demand  that  (3)  contain  the  given  point  (0,  —  4)  of  L. 
We  have,  then,  setting  x  =  0  and  y  =  —  4  in  (3) : 

(0_4-2)+fc(0  +  4)=0         or        k  =  f . 
Consequently,  the  equation  of  the  line  L  is 

x  +  y  —  2  +  |  (a;  —  y)=  0         or         5x—  y—  4  =  0. 

That  the  line  represented  by  the  latter  equation  does  actually 
go  through  the  points  (1,  1)  and  (0,  —  4)  can  be  verified 
directly. 

The  principle  which  has  been  set  forth  for  two  straight  lines 
evidently  applies  to  any  two  intersecting  curves  whatever,  so 
that  we  are  now  in  a  position  to  state  the  following  general 
theorem. 

THEOREM  1.  Let  u  =  0  and  v  =  0  be  the  equations  of  any 
two  intersecting  curves.  Then  the  equation, 

u  -f  kv  =  0,  k  =£  0, 

represents,  in  general,*  a  curve  u-hich  pa.sses  through  all  the 
points  of  intersection  of  the  two  given  curves,  and  has  no  other 
point  in  common  with  either  of  them. 

The  last  statement  in  the  theorem  is  new.  To  prove  it, 
we  have  but  to  note  that,  if  the  coordinates  of  a  point  P  satisfy 
the  equation  u  +  kv  —  0  and  also,  for  example,  v  =  0,  they 
must  satisfy  the  equation  u  =  0 ;  that  is,  if  P  is  a  point  on  the 
curve  u  +  kv  =  0,  which  lies  on  one  of  the  given  curves,  it 
lies  also  on  the  other  and  so  is  a  point  of  intersection  of  the 
two. 

*  It  may  happen  in  special  cases  that  the  locus  u  -f-  kv  =  0  reduces  to 
a  point,  as  when,  for  example, 

u  -  2  z2  +  2  y2  —  x,        v  =  x2  +  y*  —  x,       k  -  —  1. 


168  ANALYTIC   GEOMETRY 

Suppose,  now,  that  the  equations  u  =  0  and  v  =  0  represent 
two  curves  which  have  no  point  of  intersection.  It  follows, 
then,  from  the  argument  just  given,  that  the  curve 

u  +  kv  =  0,  k  =£  0, 

has  no  point  in  common  with  either  of  the  given  curves.  But 
it  may  happen,  in  this  case,  that  there  are  no  points  at  all 
whose  coordinates  satisfy  the  equation  u  +  kv  =  0.  Thus,  if 

u  =  xz  +  yz  —  1, 

v  =  x>  +  y2  -  4, 
and  k  =  —  1,  we  have 

u  -f-  kv  =  3, 

and  there  are  no  points  whose  coordinates  satisfy  the  equation 
3  =  0. 

The  general  result  can  be  stated  as 

THEOREM  2.  Let  u  =  0  and  tv  =  0  be  the  equations  of  two 
non-intersecting  curves.  Then  the  equation 

u  +  kv  =  Q,  k  =f=  0, 

represents,  in  general,  a  curve  not  meeting  either  of  the  two  given 
curves.  In  particular,  it  may  happen  that  the  equation  has  no 
locus.* 

In  the  special  case  that  u  and  v  are  linear  expressions  in 
x  and  y,  it  is  possible  to  say  more. 

Ifu  =  0  and  v  =  0  are  the  equations  of  two  parallel  straight 
lines,  the  equation 


represents,  in  general,  a  straight  line  parallel  to  the  given  lines. 
For  a  single  value  of  Jc,  the  equation  has  no  locus. 
Thus,  if  the  parallel  lines  are 


the  equation 

(5)  u  +  kv  =(1  +  fc)  x  +(1  +  k}y  +  k  =  0 

*  It  may  happen,  also,  that  the  equation  represents  just  one  point,  as 
when,  for  example, 

u  =  x2  +  j/2  —  2,         v  =  x2  +  y2  —  1.        k  =  —  2, 


CERTAIN  GENERAL  METHODS        169 

has  no  locus  when  k  —  —  1,  but  otherwise  it  represents  a  line, 
of  slope  —  1,  parallel  to  the  given  lines.  In  fact,  it  yields  all 
the  lines  of  slope  —  1,  except  the  line  v  =  0,  since,  if  we  re- 
write it  in  the  form, 


the  quantity  fc/(l  +  it)  may  be  made  to  take  on  any  value,  ex- 
cept 1,  by  suitably  choosing  k. 

Pencils  of  Curves.  All  the  lines  through  a  point,  or  all  the 
parallel  lines  with  a  given  slope,  form  what  is  called  a  pencil 
of  lines.  Equation  (5)  represents,  when  k  is  considered  as  an 
arbitrary  constant,  all  the  lines  of  slope  —  1,  except  the  line 

v  =  x  +  y  +  1  =  0; 

in  this  case,  then,  u  +  kv  =  0  and  v  =  0  together  represent  all 
the  lines  of  slope  —  1,  that  is,  a  pencil  of  parallel  lines. 

Similarly,  u  +  kv  =  0,  when  v.  =  0  and  v  =  0  are  the  lines 
(1)  and  (2),  yields  all  the  lines  through  the  point  (1,  1), 
except  the  line  (2)  ;  hence  u  •+-  kv  =  0  and  v  =  0  together 
represent  all  the  lines  through  the  point  (1,  1),  —  a  pencil  of 
intersecting  lines. 

Thus,  if  u  =  0  and  v  =  0  are  any  two  lines,  the  equations 

(6)  u  +  kv  =  0  and  v  =  0 

together  represent  a  pencil  of  lines. 

If  we  set  k  =  m/l  in  u  +  kv  =  0  and  multiply  by  I,  the  re- 
sulting equation 

(7)  he  +  mv  =  0 

is  equivalent  to  the  equation  u  +  kv  =  0  when  1  3=  0,  and  when 
I  =  0  (ra  =£  0),  it  becomes  v  =  0.  Consequently,  the  two  equa- 
tions (6)  may  be  replaced  by  the  single  equation  (7). 

The  pencil  of  lines  through  the  point  (1,  1),  for  example, 
may  now  be  given  by  the  single  equation 


where  I  and  m  have  arbitrary  values,  not  both  zero. 


170 


ANALYTIC   GEOMETRY 


In  general,  if  u  =  0  and  v  =  0  are  any  two  curves,  all  the 
curves  represented  by  the  equation 

lu  +  mv  =  0, 

where  I  and  m  have  arbitrary  values,  not  both  zero,  form  what 
is  called  a  pencil  of  curves. 

Applications.     Example  1.     Let 

u  =  x2  +  y1  +  ax  +  by  +  c  =  0, 
v  =  x*  +  y?  +  a'x  +  &'?/  +  c'  =  0, 

be  the  equations  of  any  two  circles  which  cut  each  other. 
Then  the  equation 

u  —  v=(a  —  a')'x+(b  —  W)y+(c  —  c')=0 

represents  a  curve  which  passes  through  the  two  points  of 
intersection  of  the  circles.  But  this  equation,  being  linear, 
represents  a  straight  line,  and  is,  therefore,  the  equation  of 
the  common  chord  of  the  circles. 

The  foregoing  proof  is  open  to  the  criticism  that  conceivably 

we  might  have 

a  T  a'  =  0,  *         b  -  b'  =  0, 

and  then  the  equation  u  —  v  =  0  would  not  represent  a  straight 
line.  But  in  that  case  the  circles  would  be  concentric,  and  we 

have  demanded  that  they  cut  each 

other. 

Example  2.  We  can  now  prove 
the  following  theorem  :  Given  three 
circles,  each  pair  of  which  intersect. 
Then  their  three  common  chords 
pass  through  a  point,  or  are  parallel. 

Let  two  of  the  three  given  circles 
be  those  of  Example  1,  and  let  the 
equation  of  the  third  circle  be 

w  =  x*  -f  f  +  a"x  +  V'y  +  c"  =  0. 


Mi-0 


CERTAIN  GENERAL  METHODS        171 

Then  the  equations  of  the  three  common  chords  can  be  written 
in  the  form  : 

u  —  v  =  Q,  v  —  to  =  0,  w  —  M  =  0. 

Let 

Ui  =  V  —  W,  Vi  =  W  —  U,  Wi  =  U  —  V. 

We  observe  that  the  equation, 

(8)  HI  +  Vi  -f  Wj  =  0         or          —  Wi  =  %  +  'y1? 

holds  identically  for  all  values  of  x  and  y.  Consequently,  the 
line  wv  =  0  is  the  same  line  as 

«i  +  -Wi  =  0, 

and  therefore  it  passes  through  the  point  of  intersection  of 
Ui  =  0  and  v±  =  0,  or,  if  these  lines  are  parallel,  is  parallel  to 
them.  Hence  the  theorem  is  proved. 

The  above  proof  is  a  striking  example  of  a  powerful  method 
of  Modern  Geometry  known  as  the  Method  of  Abridged  Nota- 
tion* By  means  of  this  method  many  theorems,  the  proofs  of 
which  would  otherwise  be  intricate,  or  for  whose  proof  no 
method  of  attack  is  readily  discerned,  can  be  established  with 
great  ease. 

EXERCISES 

1.    Find   the  equation  of  the  straight   line  which   passes 
through  the  origin  and  the  point  of  intersection  of  the  lines 


Ans.   I2 

2.   Find   the  equation  of   the   straight   line  which  passes 
through  the  point  (—  1,  2)  and  meets  the  lines 


at  their  point  of  intersection. 

*  The  first  general  development  of  this  method  was  given  by  the 
geometer,  Julius  Pliicker,  in  his  Analytisch-geometrische  Entwicklungen 
of  1828  and  1831. 


172  ANALYTIC  GEOMETRY 

3.  Find  the  equation  of  the  straight  line  which  passes 
through  the  point  of  intersection  of  the  lines 

5x-2y-3  =  0,        4w+  7^-11  =  0 
and  is  parallel  to  the  axis  of  y. 

4.  Find   the   equation  of  the  straight  line  which  passes 
through  the  point  of  intersection  of  the  lines  given  in  Ex.  3 
and  makes  an  angle  of  45°  with  the  axis  of  x. 

5.  Find   the  equation  of   the   straight   line  which  passes 
through  the  point  of  intersection  of  the  lines  of  Ex.  1  and  is 
perpendicular  to  the  first  of  the  lines  given  in  Ex.  3. 

Ans.   38  x  +  95^  +  58  =  0. 

6.  The  same,  if  the  line  is  to  be  parallel  instead  of  per- 
pendicular. 

7.  Find  the  equation  of  the  common  chord  of  the  parabolas 

yi  -  2y  +  x  =  0,         y*  +  2x-y  =  0. 

Ans.   x  +  y  =  Q. 

8.  The  same  for  the  parabolas 


9.   Write  the  equation  of  the  pencil  of  curves  determined 
by  the  two  curves     (a)  of  Ex.  1  ;  (&)  of  Ex.  3  ;  (c)  of  Ex.  7. 

10.   What  is  the  equation  of  the  pencil  of  circles  determined 
by  the  two  circles 

2a;—  1  =  0, 


Draw  a  figure  showing  the  pencil.     Find  the  equation  of  that 
circle  of  the  pencil  which  goes  through  the  point  (2,  4). 

11.  Find  the  equation  of  the  pencil  of  parallel  lines     (a)  of 
slope  1  ;     (&)  of  slope  —  3  ;     (c)  of  slope  AO. 

Ans.   (a)  y  =  x-\-k. 

12.  Find  the  equation  of  the  pencil  of  lines  through     (a) 
the  point  (0,  0)  ;     (6)  the  point  (3,  2)  ;     (c)  the  point  (0,  6)  ; 
(d)  the  point  (x0)  y0).  Ans.   (a)  Ix  +  my  =  0. 


CERTAIN 'GENERAL  METHODS  173 

4.   The  Equation  uv  =  0.     Consider,  for  example,  the,  equa- 
tion 

(1)  rf-jf  =  0. 

Since  x2  —  y2  =  (x  —  y}(x  -f-  y), 

it  is  clear  that  equation  (1)  will  be  satisfied 
(a)     if  (x,  y)  lies  on  the  line 

(2)  *-3f  =  0; 

(6)     if  (x,  y)  lies  on  the  line 

(3)  x  +  y  =  0; 

and  in  no  other  case.  Equation  (1),  therefore,  is  equivalent  to 
the  two  equations  (2)  and  (3)  taken  together,  and  it  represents, 
therefore,  the  two  right  lines  (2)  and  (3). 

It  is  clear  from  this  example  that  we  can  generalize  and 
say: 

THEOREM.      The  equation 

uv  =  0 
represents  those  points  (x,  y)  tvhich  lie  on  each  of  the  two  curves, 

,  u  =  0,  v  =  0, 

and  no  others. 

It  follows  as  an  immediate  consequence  of  the  theorem  that 
the  equation  OT  ...  =  ^ 

whose  left-hand  member  is  the  product  of  any  number  of 
factors,  represents  the  totality  of  curves  corresponding  to  the 
individual  factors,  when  these  are  successively  set  equal  to  zero. 

Example.     Consider  the  equation, 

x*  -  y*  =  0. 
Here,* 

X4  -  tf  =(&  -  7/2)  (cc2  +  yV)  =  (X  _  y)  (35  +  y)  (&  +  y*). 

*  It  is  true  that  the  following  equation  is  an  identity,  and  so  the  sign 
=  instead  of  =  miglit  be  expected.  The  use  of  the  sign  =  for  an  identical 
equation  is  not,  however,  considered  obligatory,  the  sign  =  being  used 
when  it  is  clear  that  the  equation  is  an  identity,  so  that  the  fact  does 
not  require  special  emphasis. 


174  ANALYTIC   GEOMETRY 

The  given  equation  is,  therefore,  equivalent  to  the  three  equa- 
x  —  y  =  0,         x  +  y  =  0,        x-  +  f-  =  0. 

The  first  two  of  these  equations  represent  right  lines.  The 
third  is  satisfied  by  the  coordinates  of  a  single  point,  the 
origin.  Since  this  point  lies  on  the  right  lines,  the  third 
equation  contributes  nothing  new  to  the  locus. 

EXERCISES 

What  are  the  loci  of  the  following  equations  ? 

/v.2         7/2 

i.  ~5"a  2-  ^+3*+2=°- 

3.  2xi  +  3xy  —  2?/2  =  0.          4.   xy  +  x  +  2y  +  2  =  0. 
5.  x-  +  xy  —  2x  —  2y  =  0.       6.   x3+xy2  =  x. 
7.  3x^y-2xy  =  Q.  8.   a4-  y*  -  2a2  +  2y2  =  0. 
9.  (x-\-  y  —  l)(x?  +  y2)=  0.     Ans.    The   line   whose   inter- 
cepts on  the  axes  are  both  1,  and  the  origin. 

10.  (z+?/)(a;"-  +  2/2  +  l)=0. 

11.  (a;  +  y)[(aj-l)2  +  ^]=0. 

12.  a^  +  x"2/  —  xy2  —  y3  =  0. 

Find,  in  each  of  the  following  exercises,  a  single  equation 
whose  locus  is  the  same  as  that  of  the  given  systems  of  equa- 
tions'. 

13.  a; -2  =  0,  2/-4  =  0. 

14.  x  =  2,  2/  =  4i 

15.  a?  +  2/-2  =  0,  x-y  +  2  =  0. 

16.  o--32/  =  5,  4a:  +  3  =  0. 

17.  *  =  £,  '^  =  _^. 
a     6'  a         b 

5.  Tangents  with  a  Given  Slope.  Discriminant  of  a  Quad- 
ratic Equation.  From  elementary  algebra  we  know  that  the 
roots  of  the  quadratic  equation 


(1) 

are 


CERTAIN  GENERAL  METHODS 

Ax2  +  Ex  +  C  =  0,  JL 


175 


2A     2A 

B        1 


From  these  formulas  the  truth  of  the  following  theorem  at 
once  becomes  apparent. 

THEOREM  1.     The  roots   of  the   quadratic   equation   (1)  are 
equal  if  and  only  if 


The  quantity  B-  —  4  AC  is  known  as  the  discriminant  of  the 
quadratic  equation  (1). 

By  means  of  the  theorem  we  shall  solve  the  following  prob- 
lem. 

Problem.  Let  it  be  required  to  find  the  equation  of  the 
tangent  to  the  parabola 

(2)  tf  =  fix, 

which  is  of  slope  -|. 

Let  L  be  a  line  of  slope  -^ 
which  meets  the  parabola  in 
two  points,  Pj  and  P2.  If  we 
allow  L  to  move  parallel  to 
itself  toward  the  tangent,  T, 
the  points  Pl  and  P2  will  move 
along  the  curve  toward  P,  the 
point  of  tangency  of  T  ;  and 

if  L  approach  T  as  its  limit,  the  points  Px  and  P2  will  approach 
the  one  point  P  as  their  limit. 

It  is  clear  that  these  considerations  are  valid  for  any  conic. 
Accordingly,  we  may  state  the  following  theorem. 

THEOREM  2.  A  line  which  meets  a  conic  intersects  it  in 
general  in  two  points.  If  these  two  points  approach  coincidence 


FIG.  5 


176  ANALYTIC   GEOMETRY 

in  a  single  point,  the  limiting  position  of  the  line  is  a  tangent  to 
the  conic.* 

In  applying  Theorem  2  to  the  problem  in  hand,  let  us  denote 
the  intercept  of  the  tangent  T  on  the  axis  of  y  by  /3.  The 
equation  of  T  is,  then, 

(3)  y  =  &  +  p. 

The  coordinates  of  the  point  P,  in  which  T  is  tangent  to  the 
parabola,  are  obtained  by  solving  equations  (2)  and  (3)  simul- 
taneously. Substituting  in  (2)  the  value  of  y  given  by  (3), 
we  have 


(4)  z2  +  4(/3  -6)  x  +  4^  =  0. 

The  roots  of  equation  (4)  are  equal,  since  they  are  both  the 
abscissa  of  P.  Accordingly,  by  Theorem  1,  the  discriminant 
of  (4)  is  zero.  Hence 

16  (£  -  6)2-  16/32  =  0,         or          -  12/8  +  36  =  0. 

Thus  ft  =  3,  and  the  tangent  to  the  parabola  (2)  whose  slope 
is  ^  has  the  equation 

(5)  x  -  2y  +  6  =  0. 

If  in  (4)  we  set  (3  =  3,  the  resulting  equation, 
a2  -12  a;  +  36  =  0, 

has  equal  roots,  as  it  should.  The  common  value  is  x  =  6,  and 
the  corresponding  value  of  y,  from  (2),  is  y  =  6.  The  coordi- 
nates of  the  point  of  tangency,  P,  are,  then,  (6,  6). 

Second  Method.  We  proceed  now  to  give  a  second  method 
of  solution  for  the  type  of  problem  just  discussed.  Let  the 
conic  be  the  ellipse 

(6)  4x2  +*/2  =  5, 
and  let  the  given  slope  be  4. 

*  A  tangent  to  a  conic  might  then  be  denned  as  the  limiting  position 
of  a  line  having  two  points  of  intersection  with  the  conic,  when  these 
points  approach  coincidence  in  a  single  point  ;  this  is  a  generalization  of 


CERTAIN  GENERAL  METHODS 


177 


It  is  evident  from  the  figure  that  there  are  two  tangents  of 
slope  4  to  the  ellipse.  Let  the  intercept  on  the  axis  of  y  of 
one  of  the  tangents  be  ft.  The  equation  of 
this  tangent  is  then 

(7)  y  =  ±x+ft. 

Our  problem  now  is  to  determine  the  value 
of  ft.  To  this  end,  let  the  coordinates  of  the 
point  of  contact  of  the  tangent  be  (x:,  y^). 
Then  a  second  equation  of  the  tangent  is, 

by  (12),  §2, 

(8)  4xjx  +  y$  =  5. 

Since  equations  (7)  and  (8),  which  we 
rewrite  as 


Fia.  6 


—  5  =  0, 

represent  the  same  line,  it  follows,  from  Ch.  II,  §  10,  Th.  5, 
that 


4  ft 

From  the  equality  of  the  first  and  third  ratios  we  have 

(9)  X!  =   -  -. 

Since  the  second  and  third  ratios  are  equal, 

£ 

fc-.jf 

Furthermore,  the  point  (a^,  y^  lies  on  the  ellipse  and  so  the 
values  of  x±  and  ylt  given  by  (9)  and  (10),  satisfy  equation  (6). 
Accordingly, 

100  ,  25      K 

— •  -\ =5,  or 

ft2       ft2  ft2 

Hence  ft  has  the  value  5  or  —  5. 

the  definition  of  §  1.  A  tangent  cannot  be  defined  as  a  line  meeting  the 
conic  in  a  single  point,  for  there  are  lines  of  this  character  which  are  not 
tangents,  viz.,  a  line  parallel  to  the  axis  of  a  parabola,  or  to  an  asymptote 
of  a  hyperbola. 


—  =1 


178  ANALYTIC   GEOMETRY 

Substituting  these  values  of  /?  in  turn  in  (7),  we  obtain 
4»  —  y  +  5  =  0,  4#  —  y  —  5  =  0, 

as  the  equations  of  the  two  tangents  of  slope  4  to  the  'ellipse 
(6).  From  equations  (9)  and  (10)  it  follows  that  the  points  of 
contact  of  these  tangents  are,  respectively,  (—  1,  1)  and 

(1,  -  1). 

Both  the  methods  described  in  this  paragraph  are  general 

in  application.  For  the  usual  type  of  problem  met  with  in  a 
first  course  in  Analytic  Geometry  either  method  may  be  used 
with  facility.  It  is,  however,  to  be  noted  that  the  second 
method  presupposes  that  the  equation  of  the  tangent  to  the 
curve  at  an  arbitrary  point  on  the  curve  is  known,  whereas 
the  first  does  not.  Accordingly,  in  case  a  curve  is  given,  for 
which  the  general  equation  of  the  tangent  is  not  known,  —  for 
example,  the  parabola,  y  =3 #2—  2#+l,  — the  first  method 
will  be  shorter  to  apply. 

EXERCISES 

Determine  in  each  of  the  following  cases  how  many  tan- 
gents there  are  to  the  given  conic  with  the  given  slope.  Find 
the  equations  of  the  tangents  and  the  coordinates  of  the  points 
of  tangency.  Use  both  methods  in  Exs.  1,  2,  3,  checking  the 
results  of  one  by  those  of  the  other. 

Conic  Slope 

1.  z2-f2/2  =  5,  2. 

Ans     !2x-y-5  =  Q>  tangent  at  (2,  -  1), 
1 2x  —  y  +  5  =  0,  tangent  at  (—  2,  1). 

2.  f-  =  3x,  f. 

3.  2*2  +  2/2  =  ll,  _|. 

4.  z2  +  87/  =  0,  2. 

5.  4x2-2/2  =  20,  3. 

6.  x*  +  f-  +  2x  =  Q,  f 

7.  6y2-5a;  =  0,  If. 


CERTAIN  GENERAL  METHODS        179 

8.  What  are  the  equations  of  the  tangents  to  the  circle 

32+2,2  =  10, 

which  are  parallel  to  the  line  3x  —  y  -\-  5  =  0? 

9.  What  is  the  equation  of  the  tangent  to  the  ellipse 

4z2  +  5?/2  =  20, 

which  is  perpendicular  to  the  line  x  +  3y  —  3  =  0  and  has  a 
positive  intercept  on  the  axis  of  y  ? 

10.  Find  the  equation  of  the  tangent  to  the  parabola 

y  =  3x-  —  2x  +  1, 

which  is  perpendicular  to  the  line  #  +  4r/  +  3  =  0. 

Ans.     4  #  —  y  —  2  =  0. 

11.  Make  clear  geometrically  that,  no  matter  what  direction 
is  chosen,  there  are  always  two  tangents  to  a  given  ellipse, 
which  have  that  direction. 

12.  How  many  tangents  are  there  to  the  parabola  y1  =  2mx, 
which  have  the  slope  0?     State  a  general   theorem   relating 
to  the  number  of  tangents  to  a  parabola  which  have  a  given 
slope. 

13.  Are  there  any  tangents  of  slope  3  to  the  hyperbola 

4aj2-y2  =  5? 
If  so,  what  are  their  equations  ? 

14.  The  preceding  exercise,  if  the  given  slope  is     (a)     1 ; 
(6)     2.     Give  reasons  for  your  answers. 

6.   General    Formulas  for    Tangents  with  a   Given    Slope. 

Consider  first  the  hyperbola 

™2  nfi 

(1)  ^-1=1. 

a2     62 

Before  attempting  to  find  a  general  formula  for  the  equations 
of  the  tangents  to  the  hyperbola,  which  have  a  given  slope,  A., 
we  shall  do  well  to  ask  if  such  tangents  exist.  In  answer  to 
this  question  we  state  the  following  theorem. 


180 


ANALYTIC   GEOMETRY 


THEOREM.     All  the  tangents  to  the  hyperbola  (1)  are  steeper 
than  the   asymptotes.     Their  slopes  X  all  satisfy  the  inequality 


(2) 


or 


Conversely,  if  X  satisfies  (2),  there  are  two  tangents  of  slope  X 

to  (1).     If,  however,  A2<62/a2,  there  are  no  tangents  of  slope  X 

to  (1). 

To  prove  the  theorem,  let  a  point  P,  starting  from  the  vertex 

A,  trace  the  upper  half  of  the  right-hand  branch  of  (1).     Then 

the  tangent,  T,  at  P,  starting  from 
the  vertical  position  at  A,  turns 
continuously  in  one  direction,  and, 
as  Precedes  indefinitely,  approaches 
the  asymptote  S  as  its  limit.  In 
other  words,  the  slope,  X,  of  T 
decreases  continuously  through  all 
positive  values  greater  than  the 
slope,  6/0,  of  S,  and  approaches  6/0 

as  its  limit.*     Consequently,  X  is  always  greater  than  b/a : 


FIG.  7 


*  The  geometrical  evidence  of  this  is  convincing,  but  not  conclusive. 
To  clinch  it,  we  give  the  following  analytical  proof  :  If  the  coordinates 
of  P  are  (x,  #),  the  slope  \  of  T  is,  by  (!!),§  2, 

X  =  —  • 

aty 

According  to  Ch.  VIII,  §  4,  eq.  (3),  -  =  ?       1     — 

y     "    I-       a2 


Hence 


When  P  traces  the  upper  half  of  the  right-hand  branch  of  (1)  and  re- 
cedes indefinitely,  x  increases  continuously  from  the  value  a  through  all 
values  greater  than  a.  Then  a2/x2  decreases  continuously  from  1  and 
approaches  0  as  its  limit ;  and  1  —  a2/x2,  and  hence  Vl  —  a2/x2,  in- 


CERTAIN  GENERAL  METHODS        181 

If  P  now  traces  the  lower  half  of  the  right-hand  branch,  X 
is  negative,  and  always  : 


These  two  inequalities  can  be  combined  into  the  single  in- 
equality (2).  Thus  (2)  is  satisfied  by  the  slope  X  of  every  tan- 
gent to  the  right-hand  branch  of  (1),  and  hence  also,  because 
of  the  symmetry  of  the  curve,  by  the  slope  X  of  every  tangent 
to  the  left-hand  branch. 

From  the  reasoning  given  in  the  first  case,  when  P  traces 
the  upper  half  of  the  right-hand  branch  of  (1),  it  follows,  not 
only  that  X  >  6/a,  but  also  that  X  takes  on  every  value  greater 
than  6/a.  Hence,  if  a  value  of  X,  greater  than  6/a,  is  arbi- 
trarily chosen,  there  is  surely  at  least  one  tangent  of  this 
slope  X  to  (1),  and  consequently,  because  of  the  symmetry  of 
the  curve,  there  are  actually  two.  Similarly,  if  a  value  of  X 
less  than  —  6/a  is  given. 

To  find  the  equations  of  the  two  tangents  of  slope  X  to  (1), 
in  the  case  that  X  does  satisfy  (2),  we  apply  the  first  of  the 
two  methods  of  §  5.  Let  the  equation  of  one  of  the  tan- 
gents be 

(3)  y  =  Xx  +  p, 

where  /@  is  to  be  determined.  Proceeding  to  solve  (1)  and  (3) 
simultaneously,  we  substitute  for  y  in  (1)  its  value  as  given  by 

(3)  and  obtain  the  equation, 

62«2  -  a2  (Xa  +  /3)2=  a2**2, 
or 

(4)  (62  -  a2X2>2  -  2  cfipte  -  a2(62  +  02)  =  0. 

The  roots  of  equation  (4)  are  both  equal  to  the  abscissa  of  the 

creases  continuously  from  0  and  approaches  1  as  its  limit.  Conse- 
quently, the  reciprocal,  1/Vl  —  a2/x2,  of  Vl  —  a2/*2  decreases  continu- 
ously through  all  positive  values  greater  than  1  and  approaches  1  as  its 
limit.  Hence,  finally,  X  decreases  continuously  through  all  positive  values 
greater  than  6/a  and  approaches  b/a  as  its  limit,  q.  e.  d. 


182  ANALYTIC  GEOMETRY 

point  of  contact  of  the  tangent  (3),  and  hence  the  discriminant 
of  (4)  must  vanish.  We  have,  then, 

4  a4£2X2  +  4  a'  (&2  +  02)  (V  -  a2X2)  =  0, 
or,  simplifying, 

(5)  '  /32  =  aW-V. 
Hence  (3  has  either  of  the  values 

±  Va2X2  -  V, 
and  the  equations  of  the  two  tangents,  written  together,  are 

(6)  y  =  Xx  ±  Va-X2  -  b\ 

Since  X  satisfies  (2),  or  the  equivalent  inequality  cCX*  —  V-  >  0, 
the  quantity  under  the  radical  is  positive  and  so  has  a  square 
root.*  We  have  thus  obtained  the  following  result. 

The  equations  of  the  tangents  to  the  hyperbola  (1),  which  have 
the  given  slope  X,  where  X  satisfies  the  inequality  (2),  are  given 
by  (6). 

Let  the  student  deduce  the  following  results,  using  either  of 
the  two  methods  of  §  5. 

The  equations  of  the  tangents  to  the  ellipse 


tchich  have  an  arbitrarily  given  slope  X,  are 
(8)  y  =  \x 


The  equation  of  the  tangent  to  the  parabola 

(9)  f-  =  2mx, 
which  has  a  given  slope  X,  not  0,  is 

(10)  y^As  +  IL. 

*  If  we  take  a  value  of  X,  for  which  X2  <  52/<z2,  then  a2X2  —  62  is 
negative  and  has  no  square  root.  Consequently,  there  are  no  tangents 
with  this  slope,  as  the  theorem  states.  Finally,  if  X  =  ±  6/a,  then 
a2\2  —  52  —  0,  and  (4)  is  not  a  quadratic  equation. 


CERTAIN  GENERAL  METHODS        183 

Condition  that  a  Line  be  Tangent  to  a  Conic.  The  two 
methods  used  to  find  the  tangent  to  a  conic  with  a  given  slope 
apply  equally  well  to  the  problem  of  determining  the  condi- 
tion that  an  arbitrary  line  be  tangent  to  a  given  conic.  In 
fact,  in  finding  the  equations  of  the  tangents  of  slope  X  to  the 
hyperbola  (1),  we  have  at  the  same  time  shown  that  the  con- 
dition that  the  line 
(11)  y  =  Xx  +  ft, 

where  we  now  consider  X  and  /3  both  arbitrary,  be  tangent  to  the 
hyperbola  (1),  is  that  X  and  ft  satisfy  the  equation  (5)  : 


Similarly,  the  work  of  deriving  formula  (8)  or  (10)  involves 
finding  the  condition  that  the  line  (11)  be  tangent  to  the  ellipse 
(7)  or  the  parabola  (9). 

Example.  Is  the  line  3x  —  2  ?/  -f-  5  =  0  tangent  to  the  hyper- 
bola x-  —  4?/2  =  4? 

It  is,  if,  when  we  write  the  equations  of  the  line  and  the 
hyperbola  in  the  forms  (11)  and  (1),  the  values  which  we 
obtain  for  X,  ft,  a-,  and  b-,  namely,  f,  4,  4,  and  1,  satisfy  (5). 
It  is  seen  that  they  do  not,  and  hence  the  line  is  not  tangent 
to  the  hyperbola. 

EXERCISES 

1.  Derive  formula  (8)  and  at  the  same  time  show  that  the 
condition  that  the  line  (11),  where  now  X  and  ft  are  both  arbi- 
trary, be  tangent  to  the  ellipse  (7)  is  that  X  and  ft  satisfy  the 
equation 

(12)  £2  =  a2X2  +  62. 

2.  Show  that  the  line  (11)  is  tangent  to  the  parabola  (9)  if 
and  only  if 

(13)  2Xft  =  m. 
Hence  prove  the  validity  of  formula  (10). 


184  ANALYTIC    GEOMETRY 

3.  By  direct  application  of  the  methods  of  the  text,  show 
that  the  condition  that  the  line  (11)  be  tangent  to  the  circle 

(14)  a2  +  f  =  a2 
is  that 

(15)  /?2  =  a2(l  +  A2). 

4.  Using  formulas  (6),  (8),  and  (10),  find  the  equations  of 
the  tangents  which  are  required  in  Exs.  1,  2,  3,  5,  and  7  of  §  5. 

5.  Has  the  hyperbola  9#2  —  4y2  =  36  any  tangents  whose 
inclination  to  the  axis  of  x  is  60°  ?     Whose  inclination  is  45°  ? 
If  so,  find  their  equations. 

6.  Find  the  equations  of   the   tangents   to   the   parabola 
yz  =  $x,  one  of  which  is  parallel  to  and  the  other  perpendicu- 
lar to  the  line  3x  —  2y  +  5  =  0.     Show  that  these  tangents 
intersect  on  the  directrix. 

7.  Prove  that  any  two  perpendicular  tangents  to  a  parabola 
intersect  on  the  directrix. 

In  each  of  the  following  exercises  determine  whether  the 
given  line  is  tangent  to  the  given  conic.  If  it  is,  find  the 
coordinates  of  the  point  of  contact. 

Conic  Line 

8.  2a?'!  +  3!  =  5,  2x  —  3-5  =  0. 


In  each  of  the  following  cases  the  equation  of  the  given  line 
contains  an  arbitrary  constant.  Find  the  value  or  values  of 
this  constant,  if  any  exist,  for  which  the  line  is  tangent  to  the 
given  conic. 

Conic  Line 

11.  z-  +  3;?/'!  =  4,  x  —  3y  +  c  =  0.     Ans.    c  =  ±  4. 

12.  ^-y2  =  3,  2x  +  dy-3  =  Q. 

13.  5y*  =  3x,  Jcx—  lOy  +  15  =  0. 

14.  4^  —  3     =  1  o:  +  2?/  +  fc  =  0. 


CERTAIN  GENERAL  METHODS 


185 


15.  Is  the  line  x  +  y  =  1  tangent  to  the  parabola  y  =  x  —  x*? 

16.  Show  that  the  lines  3x  ±  y  -f  10  =  0  are  common  tan- 
gents of  the  circle  x'-  -+•  y*  =  10  and  the  parabola  y1  =  120  x. 

17.  Find   the   equations   of   the   common  tangents  of   the 
parabola  ?/2  =  4V2a  and  the  ellipse  x2  +  2y2  =  4. 


FIG.  8 


7.  Tangents  to  a  Conic  from  an  External  Point.  Given  a 
point  P  external  to  a  conic,  that  is,  lying  on  the  convex  side  of 
the  curve.  From  P  it  is  possible,  in  general,  to  draw  two 
tangents  to  the  conic.  It  is  required 
to  find  the  equations  of  these  tan- 
gents. 

Let  the  conic  be  the  ellipse 

(1)  x2  +  2y2  =  3 

and  let  P  be  the  point  (-  1,  2).     We 

find  the  equations  of  the  two  tangents 

drawn  from  P  to  the  ellipse  by  find- 

ing   first    the    coordinates    of    the 

points  of  tangency.     Let  Pv  be  the 

point  of  tangency  of  one  of  the  tangents,  and  let  the  coordi- 

nates  of  JPj,  which  are   as   yet  unknown,  be  (xlf  y^.     The 

equation  of  this  tangent  is  then,  by  (12),  §  2, 

(2)  0^  +  2^  =  3. 

There  are  two  conditions  on  the  point  P1;  to  serve  as  a  means 
of  determining  the  values  .of  xt  and  y±.  In  the  first  place,  the 
tangent  (2)  at  PI  must  go  through  the  point  (—  1,  2)  ;  hence 

(3)  -^  +  401  =  3. 

Secondly,  the  point  Pj  lies  on  the  ellipse  (1)  ;  that  is, 

(4)  0^  +  2^  =  3. 

Equations  (3)  and  (4)  are  two  simultaneous  equations  in  the 
unknowns  xl}  yt.  If  we  solve  (3)  for  xl  : 

(5)  »1  =  4y1-3, 


186  ANALYTIC   GEOMETRY 

and  substitute  its  value  in  (4),  we  obtain,  on  simplification,  the 
following  equation  for  y± : 

(6)  3^-4^  +  1  =  0. 

The  roots  of  this  equation  are  y^  =  1  and  yl  =  ^ ;  the  corre- 
sponding values  of  Xi  are,  from  (5),  1  and  —  f .  Hence  (xlt  y^) 
=  (1,  1)  and  .(ajj,  y1)  =  (—  |,  £)  are  the  solutions  of  (3)  and  (4). 
The  coordinates  of  the  points  of  tangency  are,  therefore, 
(1,  1)  and  (— f,  ^).  Substituting  the  coordinates  of  each  point 
in  turn  for  x±  yi  in  (2)  and  simplifying  the  results,  we  obtain, 
as  the  equations  of  the  two  tangents, 

(7)  a;  +  2y-3  =  0        and        5x  -2y  +  9  =  0. 

The  method  used  in  this  example  is  universal  in  its  applica- 
tion, not  only  to  conies,  but  to  other  curves  as  well.  It  should 
be  noted,  however,  that  the  equation  corresponding  to  (6)  does 
not,  in  general,  have  rational,  that  is,  fractional  or  integral, 
roots.  Usually  its  roots  involve  radicals  and  hence  so  do  the 
final  equations  of  the  tangents.  If  one  were  dealing  with  an 
arbitrary  point  P  external  to  an  arbitrary  conic,  for  example, 
the  ellipse 


these  radicals  would  be  complicated.  Accordingly,  we  make  no 
attempt  to  set  up  general  formulas  for  the  tangents  to  a  given 
conic  from  an  external  point.  We  have  expounded  a  method 
which  is  applicable  in  all  causes,  and  this  is  the  purpose  we  set 
out  to  achieve. 

Second  Method.  We  give  briefly  an  alternative  method  of 
finding  the  equations  of  the  tangents  from  the  point  (—  1,  2)  to 
the  ellipse  (1). 

Suppose  one  of  the  tangents  is  the  line 

(8)  y  =  \x  +  (3. 

Since  it  is  a  tangent  to  (1),  we  have,  according  to  §  6,  Ex.  1, 
2^  =  6  A2 +  3. 


CERTAIN  GENERAL  METHODS  187 

Since  it  contains  the  point  (—  1,  2), 


If  we  solve  these  equations  in  X  and  ft  simultaneously,  we  find 
that  A.  =  —  i  or  |  and  that  /3  =  f  or  f  .  Substituting  these  pairs 
of  values  for  A  and  (3  in  turn  in  (8)  and  simplifying  the  results, 
we  obtain  the  equations  (7). 

EXERCISES 

1.  Make  clear  geometrically  that  from  a  point  external  to 
an  ellipse  or  a  parabola  there  can  always  be  drawn  just  two 
tangents  to  the  curve. 

2.  How  many  tangents  can  be  drawn  to  a  hyperbola  from 
its  center  ?     From  a  point  on  an  asymptote,  not  the  center  ? 
From  any  other  external  point  ?     Summarize  your  answers  in 
the  form  of  a  theorem. 

3.  Let  P  be  a  point  external  to  a  hyperbola  from  which  two 
tangents  can  be  drawn  to  the  curve.     How  must  the  position  of 
P  be  restricted,  if  the  two  tangents  are  drawn  to  the  same 
branch  of  the  hyperbola  ?     To  different  branches  ? 

4.  The  point  (2,0)  is  a  point  internal  to  the  hyperbola 
a2  —  2y2  =  2.     Prove   analytically   that    no    tangent    can    be 
drawn  from  it  to  the  curve. 

In  each  of  the  following  exercises  determine  how  many  tan- 
gents there  are  from  the  point  to  the  conic,  and  when  there  are 
tangents,  find  their  equations.  Use  the  first  method. 

Conic  Point 

9      K  /o  -i\ 

5.  **je*4  (3,1).     An,. 

6.  0^-3^  =  4,  (I,  -I)- 

7.  a?  -202  =  2,  (1,  -2). 

8.  4z2-9y2  =  36,  (4,1). 

9.  ?/2-4a;  =  0,  (4,5). 
10.  x2-4y2  =  4,  (2,1). 


188  ANALYTIC  GEOMETRY 

11.  B2-8y  =  0,  (3,2). 

12.  2z2-3y2  =  -10,  (-2,1). 

13.  x*  +  y2  -  4s  -  y  =  0,  (5,  2). 

14.  a2  +  2,2  =  25,  (-1,7). 

15.  Work  Exercises  5-10  by  the  second  method. 

16.  Show,  by  use  of  the  second  method,  that  the  tangents 
from  the  point  (2,  3)  to  the  ellipse  4^  +  9y2  =  36  are  perpen- 
dicular. 

EXERCISES  ON   CHAPTER  IX 

1,  Prove  that  the  slope  of  the  conic 

(1  —  e2)as2+  #2  —  2  mx  +  ra2  =  0 

at  the  point  fa,  y^)  is 

_  (1  —  e2)a?i  ~  m 

2/i 
Hence  show  that  the  equation  of  the  tangent  at  fa,  yj  is 

(1  —  e^x^x  +  y$  —  m(x  +  x:)  +  m2  =  0. 

2.  Show  that  the  slope  of  the  curve 

Ax*  +  Bxy  +  <7?/2  +  Dx  +  Ey  +  F=  0 
at  the  point  fa,  yj  is 


Then  prove  that  the  equation  of  the  tangent  at  fa,  y^  is 

=  0. 


3.    The   following   equations   contain  arbitrary  constants. 
What  does  each  represent  ? 
(a)  y  =  \x  +  3; 

Ans.   All  the  lines  through  (0,  3)  except  x  =  0. 


(c)  7x 

(d)  (2a 


CERTAIN  GENERAL  METHODS  189 

(e)  lx+(2l  +  m)?/-3m  =  0; 


4.  A  line  moves  so  that  the  sum  of  the  reciprocals  of  its 
intercepts  is  constant.  Show  that  it  always  passes  through  a 
fixed  point. 

5    A  line  with  positive  intercepts  moves  so  that  the  excess 

of  the  intercept  on  the  axis  of  x  over  the  intercept  on  the  axis 

of  y  is  equal  to  the  area  of  the  triangle  which  the  line  forms 

'with  the  axes.     Show  that  it  always  passes  through  a  fixed 

point. 

6.   Prove  that  the  straight  lines, 


meet  in  a  point,  by  showing  that  the  equation  of  one  of  them 
can  be  written  in  the  form  lu  +  mv  =  0,  where  u  =  0  and  v  =  0 
are  the  equations  of  the  others. 

7.  Show  that  the  three  lines, 

x  +  3y-    4  =  0, 
5x  —  3y+    6  =  0, 
3x-9y  +  U  =  0, 
meet  in  a  point. 

8.  Prove  that  the  three  lines 

Jca  —  1(3  =  0,        1(3  —  my  =  0,        my  —  Tea  =  0, 

where  cc=0,  /?=0,  and  y=0  are  themselves  equations  of 
straight  lines  and  k,  I,  and  m  are  constants,  meet  in  a  point. 

9.  Find  the  equation  of  the  common  chord  of  the  two  in- 
tersecting circles 

X2+       y2  +  Qx_8y+      3  =  0, 

2x*~  +  2y2-3x  +  4y-12  =  0. 
10.    Show  that  the  two  circles, 

a;2  +  y2  -  4z  -  4y  -  10  =  0, 
a2  -I-  y2  +  Qx  +  6y  +  10  =  0, 


190  ANALYTIC   GEOMETRY 

are  tangent  to  one  another.     Find  the  equation  of  the  common 
tangent  and  the  coordinates  of  the  common  point. 

11.  Find  the  equation  of  the  circle  which  goes  through  the 
points  of  intersection  of  the  two  circles  of  Ex.  9  and  through 
the  origin. 

12.  Find  the  equation  of  the  circle  which  is  tangent  to  the 
circles  of  Ex.  10  at  their  common  point  and  meets  the  axis  of 

x  in  the  point  x  =  2. 

• 

13.  What  is  the  equation  of  the  circle  which  passes  through 

the  points  of  intersection  of  the  line 


and  the  circle 

z2  +  2/2  +  2a-4y  +  l  =  0, 

and  goes  through  the  point  (1,  1)  ? 

14.  Determine   the   equation   of   the  ellipse  which   passes 
through  the  points  of  intersection  of  the  ellipse 

cc2  +  4  y2-  =  4 

and  the  line  3x  —  4  y  —  3  =  0, 

and  goes  through  the  point  (2,  1).  By  a  transformation  to 
parallel  axes  (cf.  Ch.  XI,  §  1),  prove  that  this  ellipse  has  axes 
parallel  to  those  of  the  given  ellipse  and  has  the  same 
eccentricity. 

15.  Find  a  single  equation  representing  both  diagonals  of 
the  rectangle  whose  center  is  at  the  origin  and  one  of  whose 
vertices  is  at  the  point  (a,  6). 

16.  What  is  the  condition  that  the  equation 

aW  -  bY  =  0 
represent  two  perpendicular  lines  ? 

17.  Find  the  locus  of  each  of  the  following  equations  : 

(a)         6a;2  +    5xy-     4y2  =  0; 
(6)         4a2 
(c)  a?' 


CERTAIN  GENERAL  METHODS        191 

18    Prove  that  the  equation 
(1)  Ax*  +  Bxy  +  Cy2  =  0 

represents  the  origin,  a  single  straight  line,  or  two  straight 
lines,  according  as  the  discriminant,  B'  —  ^AC,  is  negative, 
zero,  or  positive. 

19.  Show   that,   if   equation    (1),   Ex.    18,   represents    two 
straight  lines,  the  slopes  of  these  lines  are  the  roots  of  the 
equation 

20.  Prove  that  the  equation 

14  x2  —  45  xy  —  14  y1  =  0 
represents  two  perpendicular  straight  lines. 

21.  Show  that  equation  (1),  Ex.  18,  represents  two  perpen- 
dicular straight  lines  if  and  only  if  A  +  (7=0. 

22.  Prove  that  the  equation 

yz  —  2  xy  sec  0  +  tf  =  0 

represents  two  straight  lines  which  form  with  one  another  the 
angle  0. 

23.  A  regular  hexagon  has  its  center  at  the  origin  and  two 
vertices  on  the  axis  of  x.     Find  a  single  equation  which  repre- 
sents all  three  diagonals.  Ans.   y3  —  3x-y  =  0. 

24.  Determine  the  points  of  contact  of  the  tangents  drawn 
to  an  ellipse  from  the  points  on  the  conjugate  axis  which  are 
at  a  distance  from  the  center  equal  to  the  semi-axis  major. 

25.  Find  the  equations  of  the  common  tangents  of  each  of 
the  following  pairs  of  conies  : 

(a)         < 


/j.\         *"  i  y 1  *    i  !/' 1  . 

^  2^+Q~         '  Ifi  +  fs          J       ' 

&*J  <3  -L\J          &*j 

i\      £*._i.y!=i          ^!_^  =  i 

16      9  ~  25     16  ~ 

Draw  a  good  figure  in  each  case,  showing  the  common  tangents 


192  ANALYTIC   GEOMETRY 

26.    Show  that  the  line 
(2)  ^   "        l  +  J-1 

is  tangent  to  the  circle 


if  and  only  if  --  1  --  =  —  • 

A*     B2     a2 

27.  Find  the  condition  that  the  line  (2),  Ex.  26,  be  tangent 
to  the  ellipse 

'^=1.  Ans.    £+£  =  1. 

a2      &2  A2      W 

28.  What  will  the  condition  obtained  in  Ex.  27  become  in 
the  case  of  the  hyperbola 

—  -  y-  =  1  ? 
a2      6* 

29.  Prove   that   the   line    (2),   Ex.   26,   is   tangent    to   the 
parabola  y~  =  2  ma;,  if  and  only  if  2  JB-  +  Am  =  0. 

30.  Find  the  condition  that  the  line  y  =  \x  -f-  ft  be  tangent 
to  the  conic 

(1  —  e2)z2  +  f  —  2  mx  +  m?  =  0. 

-4ns.    03  +  mX)2  —  e2  (£2  +  m2)  =  0. 

31.  In  an  ellipse  there  is  inscribed  a  rectangle  with  sides 
parallel  to  the  axes.     In  this  rectangle  there  is   inscribed  a 
second  ellipse,  with  axes  along  the  axes  of  the  first.     Show 
that  a  line  joining  extremities  of  the  major  and  minor  axes  of 
the  first  ellipse  is  tangent  to  the  second. 


CHAPTER   X 

POLAR  COORDINATES 

1.  Definition.  It  is  possible  to  describe  completely  the 
position  of  a  point  in  a  plane  by  telling  its  distance  and  its 
direction  from  a  given  point.  This  idea  forms  the  basis  of  the 
system  of  polar  coordinates. 

Let  0  be  the  given  point,  and  draw  from  O  a  ray,  OA,  from 
which  to  measure  angles.  Let  P  be  any  point  of  the  plane. 
Denote  its  distance  from  0  by  r,  and  the 
angle  A  OP  by  6.  Then  (r,  0)  form  the 
polar  coordinates  of  the  point  P.  0  is 
called  the  pole  or  origin ;  OA,  the  prime 
direction  or  initial  ray ;  and  r,  the  radius  FIG.  1 

vector  (pi.  radii  vectores). 

When  r  and  6  are  given,  one,  and  only  one,  point  is  deter- 
mined. When,  on  the  other  hand,  a  point  is  given,  r  is  com- 
pletely determined,  but  0  may  have  any  one  of  an  infinite 
set  of  values  differing  from  one  another  by  multiples  of  360° 
(or  2?r).  Thus,  if  6'  is  one  value  of  0,  the  others  will  all  be 
comprised  in  the  formula 

0=0'±360w         (or     6' 


where  n  is  a  whole  number. 

For  the  point  0,  r  =  0 ;  but  there  is  no  more  reason  for 
assigning  to  6  one  value  rather  than  another.  As  the  coordi- 
nates of  0,  therefore,  we  take  (0,  0),  where  6  may  be  any  num- 
ber whatever. 

It  is  possible  to  define  polar  coordinates  so  that  r  can  be 
negative.  Thus  the  point  ( —  2,  30°)  would  be  obtained  by 

193 


194  ANALYTIC   GEOMETRY 

drawing  the  ray  which  makes  an  angle  of  30°  with  OA  and 
then  laying  off  on  the  opposite  ray  a  distance  of  2  units.     This 

system  of  polar  coordinates  is  not 
widely  used  in  later  work  in  mathe- 
matics,  and  even  in  analytic  geometry 

,''  A      it  is  a  matter  of  custom  rather  than 

S'T=  —2 

F  of  any  logical  necessity.     We  shall, 

therefore,    adhere    to    the    original 

definition  and  exclude  negative  values  of  r,  unless  an  explicit 
statement  to  the  contrary  is  made. 

EXERCISES 

For  use  in  these  and  later  exercises  the  student  should  pro- 
cure polar  coordinate  paper,  ruled  like  a  cobweb.     Otherwise 
he  should  use  a  scale  and  protractor. 
\Plot  the  following  points  : 
V    (1,  0°).  4.    (5,  -  30°).  ^7.    (3,  180°). 

2.    (0,1°).  *5.    (2,200°).  8.    (4,i»). 

\3.    (5,30°).  6.    (2,  -90°).  ^9.    (6,  ITT). 

10.  What  are  the  coordinates  of  the  vertices  of  a  square 
whose  center  is  at  0,  the  prime  direction  being  perpendicular 
to  a  side,  if  the  length  of  one  side  is  2  a  ? 

^11.  Write  down  the  coordinates  of  the  vertices  of  an  equi- 
lateral triangle,  the  pole  being  at  the  center  and  one  vertex 
lying  on  OA. 

12.    The  same  for  a  regular  heptagon. 

^13.   What  loci  are  represented  by  the  following  equations? 
(a)  r  =  5;         (b)  cos 6  =  0;         (c)  6  =  90°. 

2.  Circles.  Among  the  simplest  curves  in  polar  coordinates 
are 

(a)  the  circles  with  center  0.     The  equa- 
tion of  one  of  them  is 
(1)  r  =  a, 

where  a  is  the  radius.  FIG.  3 


POLAR  COORDINATES 


195 


(6)  the  circles  which  pass  through  0. 
Begin  with  one  whose  center  lies  on  OA. 
If  its  radius  is  a,  then  evidently  its  equa- 
tion  is 

r  =  2  a  cos  6. 


FIG.  4 


If  the  coordinates  of  the 

center  of  an  arbitrary  circle  through  0  are 
(a,  y),  then  the  equation  is 


A      (3) 


=  2acos(0-y). 


FIG.  5 


\  EXERCISES 

1.  Plot  directly   each   of   the   following   curves  (making  a 
convenient  numerical  choice  of  a;  as,  for  example,' 2  cm.): 

(a)         r  =  2a  sin  6 ; 
(6)         r  =  —  2  a  cos  0 ; 
c)         r  =  —  2  a  sin  0. 

2.  Obtain  each  of  the  equations  in  Ex.  1  as  a  special  case 
under  (3),  by  choosing  y  properly. 

*  3.  Circles  are  described  with  their  centers  at  the  vertices 
of  the  equilateral  triangle  of  Ex.  11,  §  1,  each  circle  passing 
through  the  center  of  the  triangle.  Find  their  equations. 

4.  A  circle  of  radius  2  has  its  center  on  OA  at  a  distance  3 
from  0.     Show  that  its  equation  is 

r2  -  6  r  cos  6  +  5  =  0. 

5.  A  circle  whose  radius  is  4  has  its  center  at  the  point 
(5,  90°).     Show  that  its  equation  is 

r2  —  10rsin0  +  9  =  0. 
6.«  Show  that  the  equation  of  any  circle  is 

r2  —  2  cr  cos  (0  —  y)  +  c2  =  p2, 

where  p  denotes  the  radius  and  (c,  y)  are  the  coordinates  of  the 
center. 

What  curve  is  represented  by  each  of  the  following  equa- 
tions ? 


196  ANALYTIC  GEOMETRY 


\8' 
\9. 


r2  —  8r  sin  0  =  9. 
10.  r2  +  2  r  cos1 0  —  2  r  sin  0  =  7. 

Nil.  r2—  2rcos#-f  2r  sin  0  =  7. 

12.  r2  —  6r  cos  0  — 8r  sin  0  =  11. 

3.   Straight  Lines.     Let  us  consider  first  a  line  L  which  does 
not  pass  through  O,  and  assume,  to  begin  with,  that  L  meets 
the  prime  direction  at  right  angles  at  the 
P  distance  h  from  0.     The  equation  of  L  is, 

evidently, 

0        h      ~~' ~A    (1)  r  cos  0  =  h. 

L 
FlQ  6  If  L  is  parallel  to  the  prime  direction 

and  at  a  distance  h  above  it,  it  is  easily 
shown  that  its  equation  is 

(2)  r  sin  0  =  h. 

Let  L,  now,  be  any  line  not  going  through  0.  Draw  a  line 
through  0  perpendicular  to  L,  and  let  B  be  the  point  in  which 
it  cuts  L  (Fig.  7).  Denote  the  length 
of  the  line-segment  OB  by  h,  and  the 
%  AOB  by  y.  Let  P :  (r,  0)  be  any 
point  of  L, 

Then  £  BOP  =  6  —  y. 

Consequently,  we  have 

(3)  rcos(0-y)  =  A  O  \A 

FIG    7 
as  the  equation  of  L. 

Bays  from  0.     The  equation  of  a  ray,  or  half-line,  emanating 
from  0,  is 

where  a  is  a  constant  angle.     Thus  0  =  0  is  the  equation  of  the 
prime  direction,  and  0  =  90°  is  the  equation  of  a  ray  drawn 


POLAR  COORDINATES  197 

from  0  at  right  angles  to  the  prime  direction.  There  are  two 
such  rays ;  the  equation  of  the  other  one  is  0  =  —  90°. 

To  the  right-hand  side  of  any  of  these  equations  can  be  added 
any  positive  or  negative  multiple  of  360°,  without  altering  the 
locus. 

Lines  through  0.  The  equation  of  the  line  through  0  per- 
pendicular to  the  prime  direction  is 

(4)  cot  6  =  0. 

For  then  0  =  90°  -or  0=-90° 

and  we  have  just  seen  that  these  are  equations  of  the  two  rays 
making  up  the  line. 

The  equation  of  any  other  line  through  0  is 

(5)  tan  0  =  c, 
where  c  is  a  constant.     Thus 

»  tan  0  =  1 

represents  a  line  through  0,  for  the  points  of  which  one  or  the 
other  of  the  two  equations 

0  =  45°  or  6  =  225° 

holds.  V 

EXERCISES 

NU.   Establish  equation  (2). 

2.   Derive  equations  (1)  and  (2)  from  equation  (3). 
"S3.   If  a  line  is  perpendicular  to  the  prime  direction  but 
does  not  cut  it,  what  is  its  equation  ?     Let  h  be  its  distance 
from  the  pole. 

4.  Find  the  equation  of  a  line  which  is  parallel  to  the  prime 
direction  and  a  distance  h  below  it. 

"\5.  Find  the  equation  of  the  line  which  cuts  the  prime 
direction  at  a  distance  of  5  units  from  the  pole  and  makes  an 
angle  of  —  45°  with  the  prime  direction. 

What  does  each  of  the  following  equations  represent? 
Make  a  plot  in  each  case. 


198 


ANALYTIC   GEOMETRY 


6.  r  cos  0  =  4. 

8.  r  cos  0  =  —  4. 

10.  r  sin  0  -+-  r  cos  0  =  3. 
12. 


>7.   r  sin  0  =  4. 
N9.   r  sin  6  =  —  4. 
Hll.   r  sin  0  —  r  cos  0  =  3. 

5r  sin0  —  12r  cos0  =  26. 


N 


17.   0  =  180°. 

w  4.  Graphs  of  Equations.  If  an  equation  in  polar  coordinates 
as  given,  which  cannot  be  reduced  to  one  of  the  forms  recog- 
nized as  representing  a  known  curve,  it  is  necessary,  in  order 
to  determine  what  curve  is  denned  by  the  equation,  to  plot  a 
reasonable  number  of  points  whose  coordinates  satisfy  the 
equation.  But  considerations  of  symmetry  will  often  shorten 
the  work. 

Example  1.     Consider  the  equation 

(1)  r2  =  16  sin  6. 
This  equation  is  equivalent  to 

(2)  r  =  4  Vsin  0, 

where  we  have  taken  only  the  positive  square  root,  since  nega- 
tive values  of  r  have  for  us  no  meaning. 

When  0  =  0,  r  =  ft;  as  0  increases,  r  increases,  and  when 
0  =  90°,  r  =  4.  Using  a  table  of  sines  and  a  table  of  square 
roots,  we  compute  the  following  coordinates  of  further  points 
of  the  curve. 

10°          20°          30°          40°          50°          60°          70°          80° 


1.67        2.34        2.83        3.21         3.50        3.72 


3.97 


—7P 


More  computations  are  unnecessary. 
For,  the  curve  is  symmetric  in  the  ray 
Q  =  90°.  To  prove  this,  we  note  that,  if 
P:  (r,  0)  is  any  point  of  the  curve,  then 
the  point  P' :  (r,  180°  -  0),  which  is 
symmetric  to  P  in  the  ray  0  =  90°,  is 
also  a  point  of  the  curve,  inasmuch  as 

sin  (180°  -  0)  =  sin  0. 


POLAR  COORDINATES 


199 


We  now  have  points  on  the  curve  for  values  of  8  from  0°  to 
180°.  These  points  determine  the  entire  curve,  since,  if  0  is 
greater  than  180°  (and  less  than  360°),  sin  0  is  negative  and  (2) 
is  meaningless. 


100°        90 


70" 


To  make  sure  that  the  curve  has  not  sharp  corners  at  0  and 
B,  we  must  compute  r  for  small  values  of  6  and  also  for  values 
of  e  near  90°. 


85° 


88° 


.53 


.92 


1.18 


3.99        4.00  - 


The  corresponding  points,  when  plotted  (Fig.  9),  show  that 
the  curve  is  smooth  at  0  and  B. 

If  we  admit  negative  r's,  we  obtain  for  90° 

each  point  of  the  present  curve  a  new  point 
symmetric  to  it  in  0.     We  have,  then,  in- 
stead  of  a  single  loop,  a  curve  with  two 
loops  (Fig.  10)  which  are  symmetric  to  each  iso2 
other  in  0,  and  also  in  OA. 

Example  2.     Given  the  equation 
(3)  r  =  10  cos  3  8. 

When  8  =  0,  r  =  10.     But  here,  as  8  in-  FIG.  10 


O 


200 


ANALYTIC   GEOMETRY 


creases,  r  decreases,  and  when  0  =  30°,  r  =  0. 
of  intermediate  points  of  the  curve  are : 


The  coordinates 


10° 


15° 


20°        25°      27.5° 


8.7 


7.1 


5.0        2.6 


1.3 


FIG.  11 


The  curve  is  symmetric  in  the  prime  direction.  For,  if  the 
point  P:  (r,  0)  is  any  point  of  the  curve, 
then  the  point  P' :  (r,  —  0),  which  is 
symmetric  to  P  in  OA  (Fig.  11),  is  also 
a  point  of  the  curve,  since 

cos3(—  0)=  cos  30. 

If  we  plot  the  points  already  computed  and  those  symmetric 
to  them,  we  obtain  a  piece  of 
the  curve  (Fig.  12).  By  plot- 
ting, further,  —  the  points  for  0 
equal,  say,  to  1°,  2°,  and  3°, — 
we  would  find  that  the  curve  is 
smooth  in  the  point  A. 

When  0  increases  beyond  0 
30°,  30  is  greater  than  90° 
and  r  becomes  negative,  so  that 
there  are  no  points  on  the 
curve.  This  situation  persists 
throughout  the  angle 

30°  <  0  <  90°. 
In  the  angle  90°  <  0  <  150°, 

however,  r  is  again  positive.  The  piece  of  the  curve  which 
lies  in  this  angle  is  congruent  to  the  piece 
OA  already  plotted  and  may  be  obtained 
by  rotating  the  piece  OA  through  an 
angle  of  120°  about  the  pole.  For,  if 
P:  (r,  0)  is  a  point  of  the  curve  (Fig.  13), 
then  P'  :(r,e  +  120°)  is  also,  since 


FIQ.  12 


O 
FIQ.  13 


cos  3  (0  +  120°)  =  cos  (30  +  360°)  =  cos  3  0. 


POLAR  COORDINATES  201 

Hence  every  point  on  the  first  lobe  of  the  curve  yields  a  point 
on  the  second  lobe  by  merely  rotating  the  radius  vector 
through  120°. 

The  second  lobe  again  gives  rise  to  a  third  lobe  congruent 
to  it  and  advanced  by  120°.  If  this  last  lobe  were  again  ad- 
vanced, it  would  yield  the  first.  Hence 
the  thTee  lobes  complete  the  curve. 

If  we  admit  negative  r's,  the  curve  is 
unchanged.     The  points  which  we  then        vv  > 

get,  for  example,  for  values  of  0  between          ^rf^r~ ^ 

30°  and  90°  lie  on  the  third  lobe  of  the       ^ 

curve.     Thus,  for  values  of  6  from  0°  to 

360°,  each  point  of  the  curve  is  obtained 

twice,    once    for    0  =  0'    and    once    for  FIQ  14 

0  =  0'  + 180°. 

Tests  for  Symmetry.  Let  the  student  show  that  the  test  for 
symmetry  in  the  ray  0  =  90°,  given  in  Example  1,  also  insures 
symmetry  in  the  ray  0  =  270°,  and  that  the  test  for  symmetry 
in  the  prime  direction,  given  in  Example  2,  also  yields  sym- 
metry in  the  ray  0  =  180°. 

These  tests  are  general,  and  can  be  stated  as  theorems. 

THEOREM  1.  A  curve  is  symmetric  in  the  line  of  the  prime 
direction  if,  on  substituting  —  0  for  0  in  its  equation,  the  equa- 
tion is  unaltered. 

THEOREM  2.  A  curve  is  symmetric  in  the  line  through  the 
pole  perpendicular  to  the  prime  direction  if,  on  substituting 
180°  -  0  for  0  in  its  equation,  the  equation  is  unaltered. 

EXERCISES 

Plot  the  following  curves. 

1.  The  lemniscate  (take  a  =  5  cm.), 

r2  =  a-  cos  2  0. 

2.  The  cardioid  (take  a  =  2-1-  cm.), 

r  =  2a(l  —  cos0). 


202  ANALYTIC   GEOMETRY 

3.  The  limagon  (a  generalization  of  the  cardioid), 

r  =  4  —  3  cos  0. 

4.  A  second  type  of  limaqon, 

r  =  3  —  4  cos  0. 

Show  that  if  negative  r's  are  admitted,  a  piece  is  added  to  the 
curve. 

5.  r2  =  16cos0.         6.  r  =  10sin30.         7.   r2  =  a-sin20. 

8.  How  are  the  curves  of  Exs.  5,  6,  and  7  related,  respec- 
tively, to  the  curves  of  Examples  1,  2  of  the  text  and  the  lem- 
niscate  of  Ex.  1  ? 

9.  r  = 10.   r  =  sec2-- 

1  +  cos  0  2 


1  -  1  cos  0  1  +  2  cos  6 

13.  r=5cos20. 

Show  that  this  curve  has  two  lobes,  but  that  it  would  have 
four  lobes,  if  negative  r's  were  admitted. 

14.  r  =  5cos40.  15.   r  =  a  cos  w0. 

16.  Show  that  the  curve  of  Ex.  15  has  n  lobes  ;  but,  if  n  is 
even  and  negative  r's  are  admitted,  it  has  2  n  lobes. 

17.  The  spiral  of  Archimedes, 

r  =  0, 

taking  6  in  degrees  and  -^  cm.  as  the  unit  of  length. 

18.  The  hyperbolic  spiral, 


taking  6  in  radians  and  2  cm.  as  the  unit  of  length. 
19.   r  =  l-  6\  20.    r  •  +  02  =  i, 

5.    Conies.     The  equation  of  a  conic  section,  when  the  defini- 
tion of  Ch.  VIII,  §  7  is  used,  is  simple  in  polar  coordinates. 


POLAR  COORDINATES 


203 


Let  the  pole  be  taken  at  the  focus  F,  and 
let  the  prime  direction  be  chosen  perpendicular 
to  the  directrix  D  and  away  from  Z);  let 
KF=m. 

If  P :  (r,  6)  is  a  point  lying  to  the  right  of  D 
(Fig.  15)  and  on  the  conic,  then 

FP=  r,  MP  —  r  cos  0  +  m. 

Now,  by  definition, 

FP^e 
MP 

and  hence  the  equation  of  the  locus  of  P  is 

r 


K 


or,  if  we  solve  for  r, 
(1) 


r  cos  0  +  m 


—  e  cos  0 


Ellipses  and  parabolas  lie  to  the  right  of  D  and  hence  are 
represented  by  (1),  when  e  <  1  and 
e  =  1,  respectively. 

When   e  >  1,    however,    (1)    repre- 
_  sents  just  the  right-hand  branch  of  a 

hyperbola.     The  equation  of  the  left- 
hand  branch  is 


Fia.  16 


(2) 


r  =  — 


em 


1  +  e  cos  0 


For,  if  P :  (r,  0)  lies  on  the  left-hand  branch  (Fig.  16),  it  is  to 
the  left  of  D  and 

PM=  —  r  cos  6  —  m. 

We  then  have,  since  FP/PM—  e, 

r 


—  r  cos  v  —  m 
which  reduces  to  (2). 

If  negative  r's  are  admitted,  the  single  formula  (1)  gives 
both  branches.     For,  in  this  case  we  may  take  r,  for  a  point 


204 


ANALYTIC   GEOMETRY 


P  on  the  left-hand  branch,  as  nega- 
tive (Fig.  17).     Then  we  have 

FP=  —  r, 


and  hence 


—  r 


FIG.  17 


—  m —  r  cos  $ 


=  e, 


which  reduces  to  (1). 

It  is  seen,  then,  that  the  choice  we  have  made,  admitting 
only  positive  or  zero  r's,  is  more  discriminating,  for  we  are 
able  to  represent  a  single  branch  of  the  curve  by  a  simple 
formula,  —  (1)  or  (2).  In  analytic  geometry  we  do  not  usu- 
ally care  to  do  this,  the  curve  that  interests  us  being  the  pair 
of  branches.  But  in  applied  mathematics  it  often  happens  that 
one  branch  of  a  hyperbola  plays  a  role  and  the  other  has  no 
meaning.  Thus  when  a  comet  is  traveling  in  a  hyperbolic  orbit, 
it  is  only  one  branch  of  the  hyperbola  which  forms  the  path.* 

New  Choice  of  Prime  Direction.  If  the  prime  direction  had 
not  been  chosen  along  KF  produced,  but  at  an  angle  y  with 
it,  as  shown  in  Fig.  18,  then  evidently 
0  —  y  would  take  the  place  of  0  in  the 
foregoing  formulas,  but  there  would  be 
no  other  change.  The  final  equations 
would  now  read 


(3) 


em 


1  —  e  cos  (0  —  y) ' 


em 


1  +  e  cos  (0  —  y) 

*  We  note  that  this  is  not  the  only  way  in  which  we  are  able,  by 
simple  formulas,  to  discriminate  between  the  two  branches  of  a  hyper- 
bola. Thus  the  equation 

-  x2  +  y2  =  1 

represents  a  hyperbola  on  the  axis  of  y.     The  equation 


y  =  Vl  +  x2 
represents  one  of  its  branches,  and  the  equation 


the  other. 


y=  —  Vl 


POLAR  COORDINATES  205 

Example.     What  curve  is  represented  by  the  equation 

6 

r  = ? 

1  +  2  sin  6 

The  equation  can  be  reduced  to  the  form  (3)  by  choosing  y 
so  that 

cos  (0  —  y)  =  —  sin  6. 

Obviously,  y  must  be  270°  or,  what  amounts  to  the  same 
thing,  —  90°.  Moreover,  e  =  2  and  m  =  3. 

Thus  the  equation  represents  one  branch  of  a  hyperbola 
whose  eccentricity  is  2.  The  transverse  axis  is  perpendicular 
to  the  prime  direction  and  the  branch  in  question  is  the  one 
opening  downward.  The  center  of  the  hyperbola  is  at  the 
point  (4,  90°).  Its  asymptotes  make  angles  of  60°  with  the 
transverse  axis.  The  lengths  of  the  semi-axes  are  a  =  2  and 
b  =  2  V3.  The  vertices  are  at  the  points  (2,  90°)  and  (6,  90°). 
The  second  focus  is  at  (8,  90°)  and  the  equations  of  the  direc- 
trices are  r  sin  0  =  3  and  r  sin  0  =  5. 

Let  the  student  verify  each  of  these  statements,  using  the 
formulas  of  Ch.  VIII,  §  6,  and  then  draw  the  curve  to  the 
scale  of  1  cm.  as  a  unit,  marking  each  of  the  points  mentioned 
with  its  coordinates  and  drawing  in  the  asymptotes  and  direc- 
trices. What  is  the  length  of  the  latus  rectum  ? 

EXERCISES 

What  conic  or,  in  the  case  of  a  hyperbola,  what  branch  is 
represented  by  each  of  the  following  equations  ?  Draw  a 
rough  figure  showing  the  position  of  the  curve. 

!.   r  = !_  2.    r=        * 


1  —  ^  cos  9  1  —  cos  6 

12  12 

3.    r  = • 4.    r  =  — 


1  —  3  cos  6  1  +  3  cos  6 

7.2  4 

5.    r  =  - — — -•  6. 


1  —  .8  sin  0  1  +  cos  0 

24  -15 

7.    r  = •  8.   r  = • 

1  +  4  sin  0  1  -  3  sin  0 


206  ANALYTIC   GEOMETRY 


9.    r  =  = = — — -^—      10.    r  = 

11.   r  = 


5  -f  3  cos  6  —  4  sin  6  1  +  sin  0  +  cos  0 

—  2 


1  +  sin  0  —  cos  0 
12.   Draw  an  accurate  figure,  to  scale,  for  each  of  the  curves 
of  Exs.  5,  6,  and  7,  marking  the  coordinates  of  all  the  impor- 
tant points  and  drawing  in  all  the  important  lines. 


6.  Transformation  to  and  from  Cartesian  Coordinates.    Let 

P  be  any  point  of  the  plane,  whose  coordinates,  referred  to  a 
pair  of  Cartesian  axes,  are  (x,  y).  Let  the 
polar  coordinates  of  P  be  (r,  6),  where 
the  origin,  0,  is  taken  as  the  pole,  and 
the  positive  axis  of  x  as  the  prime  direc- 


-x 


O\          x  tion.     Then  it  is  clear  from  the  figure  that 


(1)  x  =  r  cos  6,         y  =  rsw6. 

Thus  x  and  y  are  expressed  in  terms  of  r  and  0.     To  express 
r  and  0  in  terms  of  #  and  y,  we  have,  for  r : 

(2)  »« =  x*  +  y2,         or 
and,  for  6,  the  .pair  of  equations  : 

(3)  cos  0  =  — —  — ,  sin  B  = 


+  y2  Va;2  +  yz 

For  0  we  have,  also,  the  equation, 

(4)  tan  0  =  2. 


mi 


But  not  all  values  of  9  satisfying  this  equation  are  admissible. 
Some  determine  the  ray  OP,  as  they  should  ;  the  others  give 
the  opposite  ray  and  are  to  be  excluded.  If  0  =  0'  is  one  ad- 
missible value,  the  others  are  6  =  0'  +  360  n,  where  n  is  a 
whole  number. 

Example  1.     What  are  the  polar  coordinates  of  the  point 
(-5,  -5)? 


POLAR  COORDINATES  207 


Here,  r  =  V52  +  o2  =  5  V2,         tan  $  =  1. 

But  0  =  45°  is  not  a  correct  value  of  0,  for  the  point  lies  in  the 
third  quadrant.     The  values  of  6  are,  then  : 

6  =  225°  +  360°  n. 

It  is  frequently  of  importance  to  obtain,  from  the  equation 
of  a  curve  in  one  system  of  coordinates,  its  equation  in  the 
other  system.  We  illustrate  the  method  of  doing  this  by  a 
number  of  examples. 

Example  2.     Find  the  equation  of  the  equilateral  hyperbola 


in  polar  coordinates, 

Replacing  x  and  y  by  their  values  as  given  by  (1),  we  have  : 

r2  cos2  B  —  r2  sin2  0  =  a2, 
or  r2  cos  2  6  =  a-. 

Example  3.     Transform  the  equation  of  the  lemniscate, 
r2  =  a2  cos  2  8, 

to  rectangular  coordinates. 

We  perform  the  transformation  piecemeal,  first  getting  rid 
of  0.     Write  the  equation  as 

r2  =  a2(cos20  -  sin20), 

and  then  replace  cos  0  and  sin  0  by  their  values,  x/r  and  y/r, 
from  (1)  ;  on  multiplying  both  sides  of  the  resulting  equation 

by  r2,  we  have 

r4  =  a2  (a;2  —  y2). 

Finally,  we  replace  ?-2  by  its  value  x2  +  y2,  and  obtain 

(x2  +  2/2)2  =  a2(x2-y2). 
This  is  an  equation  of  the  fourth  degree  in  x  and  y. 

Example  4.     Transform  the  equation  of  the  curve  of  Exam- 
ple 1,  §  4 

(5)  r2  =  a2  sin  0 

to  rectangular  coordinates. 


208  ANALYTIC   GEOMETRY 

i 

Replacing  sin  6  by  y/r  and  multiplying  through  by  r,  we 
have 

(6)  7*=a*y. 
This  becomes 

(7)  (  V»2  +  ?/2)3  =  azy,        or         V(ar2  +  y2)3  =  a2y. 

Negative  Values  of  r.     If  we  admit  negative  values  of  r,  then 
(2)  and  (3)  become 

r  =  ±  Vfl?2  -f-  y2, 


cos0  = 


±  V  a?  +  .V'2  ±  V 

where   the   plus  signs  are  to  be   taken  if   r  is   positive,  the 
minus  signs  if  r  is  negative  ;  (1)  does 
not   change,  as   Fig.   20   shows.     The 
x    admissible  solutions  for  0  of  (4)  are 
those  determining   the  ray   OP  •  (Fig. 
19),  if  r  is   positive,  or  those   deter- 
mining  the  ray  OP  (Fig.  20),  if  r  is 
negative. 
If   in  (5)  negative  values  of  r  are  admitted,  then  (6)   be- 

comes, since  now  r  =  ±  Vo;2  +  y2, 


which  may  be  written  as 
(8)  ( 

The  fact  that  (5)  transforms  into  (7)  when  negative  r's  are 
excluded  and  transforms  into  (8)  when  negative  r's  are  ad- 
mitted corresponds  to  the  fact  that  in  the  first  case  the  curve 
(5)  consists  of  a  single  loop,  whereas  in  the  second  it  is  made 
up  of  two  loops  (§4). 

This  situation  is  not  .met  with  in  the  case  of  the  lemniscate, 
Example  3,  since  this  curve  is  the  same  whether  negative  r's 
are  excluded  or  admitted. 


POLAR  COORDINATES  209 

EXERCISES 

1.  Find  the  Cartesian  coordinates  of  the  points  : 
(a)     (2,60°);         (6)     (5,120°);         (c)     (10,225°); 
(d)     (3.281,  110°  32');         (e]     (2.847,  242°  27'). 

Plot  the  given  point  each  time  and  check  the  results  by  direct 
measurement. 

2.  Find  the  polar  coordinates  of  the  points  : 

(a)     (6,6);  (6)     (-  2,  -  2)  ;         (c)     (2,3); 

(d)     (-4,3);          (e)     (7,  -8);  (/)     (-  12,  -  5). 

Transform  the  following  equations  to  polar  coordinates. 

3.  x  =  3.  4.   y  =  —  4. 

5.   y  =  3x.  6.   2x  —  3y  =  8. 

7.    y2  =  4x.  8.    xy  =  a?. 

9.   z2  +  ?/2  —  2a  +  4y=0.         10.   4x2  +  3?/2=12. 

ll^Tr^asform  the  equation  of  the  cardioid, 

r  =  2a(l  —  cos-0), 

esian   coordinates.     Of   what   degree  is  the  resulting 
on  ? 
Ans.    (x2  +  ^  +  2  ax)2  =  4  a2  (x2  +  y~),  of  the  fourth  degree. 

Find  the  equation  in  Cartesian  coordinates  of  each  of  the 
following  curves. 

12.   r2=a2cos0.  13.   7-"  =  a2  sin  20. 

14.   r  =  a  esc  6.  15.    r  =  4  —  3  cos  0. 

16.  r  =  a  cos  3  0.  Ans.    (x2  +  y2)2  =  ax  (xz  —  3  ?/2). 

17.  r  =  a  sin  3  6.  Ans.    (xz  +  y2*)2  =  ay  (3  x2  —  y2). 

18.  r  =  a  sin  2  0.  Ans.    V(»2  +  y2)3  =  2  axy. 

19.  ?•  =  a  sin  2  0,  if  negative  r's  are  admitted. 

Ans.    (x2  +  y2)3  =  4  a2a;22/2. 

20.  r  =  a  cos  2  0.  Ans.    V(a;2  +  y2)3=  a  (a2  —  y2). 

^ 

21.  r  =  a  esc2  —  Ans.    y2  —  4aa:  —  4  a2  =  0. 


A  £Vi  equati 
Ans 


210  ANALYTIC   GEOMETRY 

OO        *)*  — • if   g    "^^    1 

1  —  e  cos  0 ' 

Ans.  (1  —  e2)  «2  +  y2  —  2  e2m»  —  e2m2  =  0. 


23.  r  = — ,  if  e  >  1.  ^4ws.    Vo^+1/2  =  e(*  +  m). 

1  —  e  cos  0 

24.  Same  as  Ex.  23,  if  negative  r's  are  admitted. 

Ans.   That  to  Ex.  22. 

EXERCISES  ON  CHAPTER  X 

1.    Show  that  the  distance  between  the  two  points  (rt,  6^, 
(r2,  02)  is  given  by  the  formula 


D  =  Vrx2  +  r22  —  2  rjr2  cos  (0j  -  02). 

2.  Deduce  a  formula  giving  the  area  of  a  triangle,  one  of 
whose  vertices  is  at  the  pole. 

3.  Determine  the  angle  of  intersection  of  the  two  lines 

r(sin  $  +  cos  6)=  3,         r(4  sin  0  +  3  cos  0)=  5. 
Suggestion.     Put  the  equations  into  the  normal  form, 

rcos(0  —  y)=  h, 
and  thus  find  the  value  of  y  for  each  line. 

4.  Show  that  the  line, 

s+f-i, 

a     6 
is  represented  in  polar  coordinates  by  the  equation 

ab 

r  =  -- 

a  sin  0  +  6  cos  0 

5.  What  lines  are  represented  by  the  following  equations  ? 
Plot  the  line  each  time. 


-  _  '  _  _  • 

sin  0  +  cos  0'  2  sin  0  —  3  cos  0 

Find   the   equations  in  polar  coordinates  of   the  following 
conies. 

6.     z  =  2mx.  Ans.   r  =  2m  cos  0  esc2  0. 


POLAR  COORDINATES  211 


7.       +     =!.  Ans.  r*  = 


.  . 

a2     d2  a2sin20 


, 
a2     62  -a2sin20'+62cos20 

9.   What  curves  are  represented  by  the  following  equations  ? 


400 
~ 


25  sin2  0  +  16  cos2  0  cos2  0  -  sin2  0 


2cos20+3sin20'  cos2 0 -  2  sin2 0 ' 

(e\    r  =  4cos0.  ,,,    ?.  =  4sin0 

sin2  0  '  cos2  0 

10.    Transform  the  equation  of  the  circle, 


to  polar  coordinates;  represent  the  polar  coordinates  of  the 
center  by  (c,  y).  Ans.    r2  —  2  cr  cos  (0  —  y)  +  c2  =  p2. 

CONICS 

11.  A  comet  moves  in  a  parabolic  orbit  with  the  sun  as 
focus.     When  the  comet  is  40,000,000  miles  from  the  sun,  the 
line  from  the  sun  to  it  makes  an  angle  of  60°  with  the  axis  of 
the  orbit  (drawn  in  the  direction  in  which  the  curve  opens). 
How  near  does  the  comet  come  to  the  sun  ? 

12.  A  comet  is  observed  at  two  points  of  its  parabolic  orbit. 
The  focal  radii  of  these  points,  neither  of  which  is  the  vertex 
of  the  parabola,  make  an  angle  of  90°  with  one  another  and 
have  lengths  of  10,000,000  and  20,000,000  miles,  respectively. 
Find  the  equation  of  the  orbit  and  determine  how  near  the 
comet  comes  to  the  sun. 

13.  An  ellipse  which  has  a  focus  at  the  pole  and  its  trans- 
verse axis  along  the  prime  direction  passes  through  the  two 
points  (4,  60°)  and  (2,  90°).     What  is  its  equation?     Where  is 
the  second  focus? 

14.  A  hyperbola  has  its  transverse  axis  along  the  prime 
direction  and  a  focus  in  the  pole.     The  branch  adjacent  to  this 


212  ANALYTIC   GEOMETRY 

focus  goes  through  the  points  (|V2,  45°),  ( V2,  90°).     Find  the 
equation  of  this  branch. 

15.  Show  that  in  a  parabola  a,  focal  radius  inclined  at  an 
angle  of  60°  with  the  direction  in  which  the  curve  opens  is 
equal  in  length  to  the  latus  rectum. 

16.  Show  that  a  focal  radius  of  a  hyperbola  which  is  parallel 
to  an  asymptote  is  equal  in  length  to  a  quarter  of  the  latus 
rectum. 

17.  Prove  that  in  any  conic  the  sum  of  the  reciprocals  of 
the  segments  of  a  focal  chord  is  constant. 

18.  Prove  that  the  length  of  the  focal  chord  of  any  conic  is 
given  by  the  formula 

2  em 


l-e2cos200' 

where  00  is  the  angle  which  the  chord  makes  with  the  trans- 
verse axis. 

19.  Show  that  the  sum  of  the  reciprocals  of  the  lengths  of 
two  perpendicular  focal  chords  of  a  conic  is  constant. 

ROTATION  OF  THE  PRIME  DIRECTION 

20.  Let  the  prime  direction  OA  be  rotated  about  0  through 
the  angle  00  into  a  new  position  OA'.     Let  an  arbitrary  point 
have  the  coordinates  (r,  8)  with  respect  to  0  as  pole  and  OA  as 
prime  direction  and  the  coordinates  (/,  0')  with  respect  to  0  as 
pole  and  OA'  as  prime  direction.     Show  that 

r'  =  r,  O'  =  0-  eo. 

21.  Equation  (1),  §  5,  when  e  <  1,  represents  an  ellipse  with 
one  focus  in  0  and  the  other  on  OA.     From  it  obtain,  by  rota- 
tion of  the  prime  direction,  the  equation  of  an  ellipse  with  one 
focus  at  0  and  the  other  on  the  ray  0  =  90°. 

22.  Equations   (1)  and  (2),    §  5,  when   e  >  1,   represent  a 
hyperbola  with  one  focus  in  O  and  the  other  on  the  ray  0=180° ; 
obtain  from  them  the  equations  of  a  hyperbola  with  one  focus 
at  0  and  the  other  on  the  ray  6  =  90°. 


POLAR  COORDINATES  213 

23.  Obtain  equations  (3)  and  (4),  §  5,  from  equations  (1) 
and  (2),  §  5,  by  a  rotation  of  the  prime  direction. 

24.  What  does  the  equation  of  the  line 

r(sin  0  +  cos  0}  =  3 

become,  when  it  is  referred  to  the  perpendicular  to  it  from  the 
pole  as  the  new  prime  direction  ? 

25.  By  rotating  the  prime  direction  through  a  suitable  angle 
reduce  the  equation  of  the  circle,  r  =  6  cos  (0  —  30°),  to  simpler 
form. 

26.  The  same  for  the  circle,  r  =  4  cos  0  +  3  sin  0. 

27.  The  same  for  each  of  the  conies : 

i  \  ^  fh\  1 

3 -3  cos  0  —  4  sin  0'  V2  +  sin  0  -  cos  0 

28.  Prove  that  the  curves  r  =  a  sin  2  0  and  r  =  a  cos  2  0  are 
the  same  curves,  referred  to  a  common  pole  and  to  prime  direc- 
tions making  an  angle  of  45°  with  one  another. 

29.  Show  that  the  equations  r  =  a  cos  3  0  and  r  =  a  sin  3  0 
represent  the  same  curve. 

30.  Show  that  the  equation  of  the  curve  r  =  a  cos  3  0  remains 
unchanged  if  the  prime  direction  is  turned  through  any  angle 
which  is  an  integral  multiple  of  120°. 

31.  The  same  for  the  curve  r  =  a  sin  40,  if  the  angle  is  90°. 

POLE  IN  AN  AKBITBABT  POINT 

32.  Given  the  point  (2,  3)  in  the  Cartesian  plane.     The  polar 
coordinates  of  a  point  P,  referred  to  (2,  3)  as  pole  and  to  the 
directed  line  through  (2,  3)  in  the  direction  of  the  positive  axis 
of  x  as  prime  direction,  are  known  to  be  (2,  13°).     What  are 
the  rectangular  coordinates  of  P  ? 

Ans.     (2  +  2  cos  13°,  3  +  2  sin  13°) 

33.  The  polar  coordinates  of  a  point  P,  referred  to  the  point 
(xw  2/o)  as  P°le  an^  the  directed  line  through  (xQ,  ?/„)  in  the 


214  ANALYTIC   GEOMETRY 

direction  of   the  positive  axis   of  x  as  prime  direction,  are 
(r,  6}.     Show  that  the  rectangular  coordinates,  (x,  y),  of  Pare 

x  =  x0  -f-  r  cos  6,        y  =  y0  •+•  r  sin  0. 

34.  By  the  direction  6  is  meant  the  direction  which  the  posi- 
tive axis  of  x  would  assume  if  it  were  rotated  about  one  of 
its  points  (in  the  positive  sense  of  rotation)  through  the  angle  6. 

From  the  point  (5,  2)  one  proceeds  2  units  distance  in  the 
direction  135°,  and  from  the  point  thus  reached  one  proceeds  3 
units  distance  in  the  direction  60°.  What  are  the  coordinates 
of  the  final  position? 

35.  Prove  that  the  equations  of  Ex.  33  can  be  considered  as 
the  equations  of  a  transformation  of  coordinates,  from  (a;,  y)  to 
(r,  0),  which  consists  first  of  a  change  of  origin  to  the  point 
(x0,  yQ)  — :  cf .  Ch.  XI,  §  1  —  and  then  of  the  introduction  of  polar 
coordinates. 

36.  By  shifting  the  origin  to  the  point  (2,  1)  and  then  in- 
troducing polar  coordinates,  identify  the  locus  of  the  equation 

y^-2x-  2y  +  4  =  0. 

37.  By  shifting  the  origin  to  a  suitable  point  and  then  in- 
troducing polar  coordinates,  identify  the  locus  of  the  equation 

(x*  +  y2  -  2x  -  2y  +  2)2  =  25  (a2  -  f-  -2x  +  2y). 

Ans.   A  lemniscate. 

Loci 

Solve  the  following  problems  in  loci,  using  polar 
coordinates  and  excluding  negative  r's.  In  Exs. 
39^41,  determine  when  two  equations  are  necessary  '  •* 

to  represent  the  locus. 

38.  Cissoid  of  Diodes.     OA  is  a  fixed  diameter 
of  a  circle.     A  variable  secant  through  0  meets 
the  circle  in  M  and  the  tangent  at  A  in  N.     De- 
termine the  locus  of  the  point  P,  so  situated  on  the 
segment  ON  that  OP  =  MN,  and  plot  it. 

39.  LimaQon  of  Pascal.     A  variable  secant  through  a  fixed 


POLAR  COORDINATES 


215 


point  0  of  a  circle  of  diameter  a  meets  the  circle  again  in  R. 
The  constant  length  b  is  laid  off  in  both  directions  along  the 


secant  from  K.  Find  the  locus  of  the  two  points  thus  reached. 
Show  that,  if  a  =  6,  the  locus  is  a  cardioid.  Plot  the  locus 
(a)  when  a  =  4,  b  =  5  ;  (6)  when  a  =  4,  b  =  3.  Of.  §  4,  Exs.  2, 
3,4. 

40.  *  Conchoid  of  Nicomedes. 
A  variable  straight  line  through 
a  fixed  point  0  meets  a  fixed 
straight  line,  at  the  distance  a 
from  0,   in  Q.     From    Q  the 
constant   length  b   is  laid  off 
in  both  directions   along   QO. 
Find    the    locus    of    the    two 
points  thus  reached.     Plot  it 

for  each  of  the  following  pairs  of  values  of  a  and  b  : 
a  =  4,  6  =  6;         a  =  6  =  4 ;         a  =  4,  6  =  3. 

41.  Ovals  of  Cassini.     Given  two  points,  Fl  and  F2,  with 
coordinates  (a,  0°),  (a,  180°).     Determine  the  locus  of  a  point 


b=a 


b<a 


©     G 


b>a 


P  which  moves  so  that  the  product  of  its  distances  from  Fl 
and  F2  is  constant,  and  equal  to  6%  Show  that,  if  a2  =  62,  the 
locus  is  a  lemniscate.  Plot  the  locus  (a)  when  a  =  6,  6  =  7; 
(6)  when  a  =  6,  6  =  5. 


CHAPTER   XI 


lAN^FORMATION  OF  COORDINATES 


1.   Paralle  Axes.     It  sometimes  happens  that  it  is  desirable 
to   shift   from  a  given   system  of   Cartesian  axes  to  a  new 

system  of  axes  having  the  same  direc- 
tions as  the  old,  but  with  a  different 
origin. 

Let  P  be  any  point  of  the  plane ; 
let  the  coordinates  of  P,  referred  to 
the  old  axes,  be  (x,  y),  and  let  the 
coordinates  be  (a/,  y')  with  respect  to 
the  new  axes.  Let  the  new  origin, 

0',  have  the  coordinates  (x0,  yQ)  in  the  old  system.     Then  it  is 
easy  to  show  that 

(1) 


V 

P 

b 

O' 

X9 

0 

ff.- 

v^ 

FIG.  1 


=  x 


x0, 


or 

(2) 


x'  —  x  —  x0, 


For,  consider  the  line-segment  OP  and  the  broken  line,  00' P, 
which  has  the  same  extremities.  Then 

Proj.  OP  =  Proj.  00'  +  Proj.  O'P, 

no  matter  what  direction  is  chosen,  along  which  the  projection 
is  to  take  place  (Introduction,  §  3). 

If  the  direction  is  taken,  first,  as  the  positive  axis  of  x  and 
then,  again,  as  the  positive  axis  of  y,  we  obtain,  by  applying 
the  definition  of  coordinates  (Ch.  I,  §  1),  the  equations  (1). 

Example  1.     Find  the  equation  of  the  curve 

216 


TRANSFORMATION   OF   COORDINATES 


217 


referred  to  parallel  axes,  with  the  new  origin 
at  the  point  (2,  —  1). 

Here,        XQ  =  +  2,        y0  =  —  1, 
and  we  have 

(4)  aj  =  aj'  +  2 

Hence 


y  =  y'  - 


FIG.2 


or,  on  simplification, 


Thus  the  curve  is  seen  to  be  a  parabola  whose  vertex  is  at 
the  new  origin.  Referred  to  the  old  axes,  the  vertex  is  at  the 
point  (2,  —  1)  and  the  focus  at  (3,  —  1). 

Example  2.     What  curve  is  represented  by  the  equation 

9x*  +  4:f  +  ISx—  16  y=  11? 
We  can  rewrite  the  equation  in  the  form  : 


The  first  parenthesis  becomes  a  perfect  square  if  1  is  added. 
This  means  that  9  must  be  added  to  each  side  of  the  equation. 
Again,  the  second  parenthesis  becomes  a  perfect  square  if  4 
is  added.  This  means  that  16  must  be  added  to  each  side  of 
the  equation.  Hence,  finally, 


or  9 

If  we  transform  to  parallel  axes,  setting 

f  X'  =  X  +  1,  XQ  =  —  1, 

\y'  =  y-2,  y0  =  2, 

the  equation  becomes 


or 


218 


ANALYTIC   GEOMETRY 


This  equation  represents  an  ellipse  with  its  center  at  the 
new  origin,  (x0,  y0)  =  ( —  1,  2) ;  its  semi-axes  are  of  lengths  2 
and  3  and  its  foci  lie  on  the  y'-axis  at  the 
points 

(*',y')=(<>,  ±V5), 
or  (a,  y)  =  (— 1,  2  ±V5). 

If,  in  Example  1,  the  position  of  the 
new  origin  (the  vertex  of  the  parabola) 
had  not  been  given,  it  could  have  been 
found  by  the  method  employed  in  Ex- 
ample 2.  Equation  (3)  can  be  written  as 

To  complete  the  square  of  y*  -f  2  y,  add  1  to  each  side  of  the 
equation : 


FIG.  3 


Put  this  into  the  form  : 


Hence  we  are  led  to  set 

x'  =  x  —  2,         y'  =  y  +  l, 

that  is,  to  transform  to  parallel  axes  with  the  new  origin  at 
the  point  (2,  —  1).  But  this  is  precisely  the  transformation 
(4)  which  was  applied  in  Example  1. 

/  EXERCISES 

Vl.  Find  the  coordinates  of  the  points  (3,  2),  (—2,  5), 
(—  4,  —  1),  (0,  0),  referred  to  new  axes  having  the  same  direc- 
tions as  the  old,  if  the  new  origin  is  at  the  point 

(«)     K  2/o)  =  (1,1);  (6)     fa,  2/o)  =  (5,  -3). 

In  each  of  the  following  exercises  transform  the  given  equa- 
tion to  parallel  axes  having  the  same  directions,  the  new  origin 
being  at  the  point  specified.  Thus  identify  the  curve  repre- 
sented by  the  equation,  and  describe  carefully  its  position  with 
respect  to  the  original  axes.  Draw  the  curve  roughly. 


TRANSFORMATION   OF  COORDINATES  219 

Equation  New  Origin 

=  2z  +  4,  (x0,y0)  =  (-2,0). 

Ans.   A  parabola  with  its  vertex  at  (—2,  0),  with  its  focus 
at  (•+-  -|  ,  0),  and  with  x  =  —  f  as  its  directrix. 

K,2/o)  =  (-i,  1). 
=  0,  (ajb,  </0)  =  (l,  2). 

18a;  -  50  y  -  191  =  0,  (z0,  y0)  =  (-  1,  !)• 

ins.    An   ellipse,    center   at    (—  1,   1);  foci   at   (3,  1)  and 
(  —  5,  1)  ;  semi-axes  of  lengths  5  and  3. 

Ify  x*  -4:f-6x-32y-59  =  0,  (xa,  y0)  =  (3,  -  4). 


Show  that  each  of  the  following  equations  represents  a  conic 
^Section.     Draw  a  rough  graph  of   the   conic  and   find,  when 
they  exist,  the   coordinates  of   the  center   and   the  foci,  the 
equations  of  the  directrices  and  the  asymptotes,  and  the  value 
of  the  eccentricity.* 


8. 

9.   y--12a;  +  4y  +  28  =  0. 

m)  2ic2  +  3y2-4a:-62/-l  =  0. 

12.  o;2  +  4?/2+2aj — 24?/  +  36  =  0. 

13.  4x2  —  9?/2  —  16  x  +  18  y  —  29  =  0. 

Ans.  A  hyperbola,  center  at  (2,  1) ;  foci  at  (2  ±  Vl3,  1) ; 
directrices  :  x  =  2  ±  T9^\/13 ;  asymptotes :  2  x  +  3y  —  1  =  0, 
2a;-3y— 1  =  0;  e  =  £V13. 


15. 

16.   a2-25i/2-  50^-50  =  0. 


2.   Eolation  of  the  Axes.     Let  the  new,_£»'7*1/')-axes   have 
the  same   origin   as   the   given  (x,  ?/)-axes,  and  let  the  angle 

*  Further  exercises  of  this  type  are  Exs.  1-9,  of  Ch.  XII,  §  1. 


220  ANALYTIC   GEOMETRY 

from  the  positive  axis  of  x  to  the  positive  axis  of  x'  be  de- 
noted by  y  (Fig.  4).  Let  P  be  any  point,  whose  coordinates, 
referred  to  the  old  and  the  new  axes, 
are  (x}  y)  and  (x',  y1)  respectively,  and 
let  M  and  M'  be  the  projections  of  P 
on  the  axes  of  x  and  x'  respectively. 
Then  0  is  joined  with  P  by  two  broken 
lines,  namely,  OMP  and  OM'P.  It 
follows  that  the  projections  of  these 
broken  lines  along  any  direction  are  equal : 

(1)         Proj .  0 M  +  Proj .  MP  =  Proj .  0 M'  +  Proj .  M'P. 

If  the  direction  is  taken,  first,  as  the  positive  axis  of  x  and 
then,  again,  as  the  positive  axis  of  y,  we  have 

OM  =  OM1  cos  y  -  M'Psiu  y, 
MP  =  OM1  sin  y  +  M'P  cos  y. 

But,  by  the  definition  of  coordinates  (Ch.  I,  §  1), 

OM=x,         MP  =  y;  OM'  =  x',         M1P  =  y'. 

The  final  result  is,  then,  the  following : 

x  =  x'  cos  y  —  y'  sin  y, 
y  =  x'  sin  y  +  y'  cos  y. 

To  express  x'  and  y'  in  terms  of  x  and  y,  these  equations  can 
readily  be  solved  for  the  former  variables,  regarded  as  the  un- 
known quantities  in  the  pair  of  simultaneous  equations  (2). 
Or,  the  formulas  can  be  deduced  directly  from  the  figure  by 
taking  the  projections  in  equation  (1)  along  the  positive  axes 
of  x'  and  y'  in  turn.  The  result  in  either  case  is 


(3) 


x'  =      x  cos  y  -f  y  sin  y, 
y'  =  —  x  sin  y  +  y  cos  y. 


Example   ~L     Transform   the   equation   of    the    equilateral 
hyperbola, 

x'i  —    i  =  a2 


TRANSFORMATION  OF  COORDINATES 


221 


to  new  axes  with  the  same  origin,  the  angle  from  the  positive 
axis  of  x  to  the  positive  axis  of  x'  being  —  45°. 
Here,  y  =  —  45°,  and  formulas  (2)  become 


V2 
=  ±(x'*3T2x'y<  +  y'*), 


Hence 


On  substituting  these  values  in  the  given 
equation  we  have  : 

2  x'y'  =  a2. 


FIG.  5 


This,  then,  is  the  equation  of  an  equilateral  hyperbola  re- 
ferred to  its  asymptotes  as  the  coordinate  axes. 

If  we  had  rotated  the  axes  through  +  45°  instead  of  —  45°, 
the  transformed  equation  would  have  read : 


=  —  of. 
Example  2.     Transform  the  equation 


or 


B*     A2 


to  new  axes,  obtained  by  rotating  the  given  axes  about  the 
origin  through  90°. 

Here,  y  =  90°,  and  equations  (2)  become 


Hence 


A* 


Thus  it  appears  that  the  original  equation  represents  a 
hyperbola  with  its  center  at  the  origin,  its  transverse  axis 
lying  along  the  axis  of  y  ;  cf  .  Ch.  VIII,  §  8.  The  length  of  the 
major  axis  is  2B,  that  of  the  minor  axis,  2  A.  The  asymptotes 
are  given  by  the  equations 

x'      y' 


and 


— 

B     A 


222  ANALYTIC   GEOMETRY 

Referred  to  the  original  axes,  they  have  the  equations 

—  =  y~         and         —  =  _2L. 
A     B  A         E 

EXERCISES 


Obtain  the  equations  of  transformation  in  each  of  the  fol- 
win 

1. 


lowing  three  cases : 

When  y  =  30°.  Ans. 

"'  =  t1 

2.    When  y  =  -  120°.  3.   When  y  =  135°. 

Draw  a  figure  and  deduce  from  it  directly  the  formulas  of 
transformation  in  each  of  the  following  three  cases.  Check 
the  results  by  application  of  formulas  (2)  and  (3)  of  the 
text. 

4.     y  =  90°.  5.     y=-90°.  6.    y=180°. 

7.  Find  the  coordinates  of  the  points  (2,  0),  (3,  1),  (—  2, 4), 
(—  5,  —  8),  referred  to  new  axes  obtained  by  rotating  the  old 
through  an  angle  of  45°  ;  of  150°. 

8.  Show  directly  by  means  of  a  suitable  rotation  of   the 
axes  that  the  equation  xy  =  k2  represents  an  equilateral  hyper- 
bola referred  to  its  asymptotes  as  the  coordinate  axes.     Deter- 
mine the  coordinates,  referred  to  these  axes,  of  the  vertices 
and  the  foci. 

9.  The  same  for  the  equation  xy  =  —  k2. 

By  rotating  the  axes  through  an  angle  of  45°  determine  the 
curve  represented  by  each  of  the  equations : 

N^A.    17x*-l6xy  +  172/2  =  225. 

11.  3x°  —  lOxy  +  3y2  +  8  =  0. 

12.  z2  -t-lxy  +  y*  +  3  =  0. 

13.  By  rotating  the  axes  through  30°  determine  the  curve 
represented  by  the  equation 


TRANSFORMATION  OF  COORDINATES 


223 


«.\  By  rotating  the  axes  through  an  angle  y  of  the  first 
quadrant,  whose  sine  is  f ,  determine  the  curve  represented  by 
the  equation 

52 a;'- 72 ay +  73 y2  =  100. 

15.  Show  that  the  equation  of  a  circle  whose  center  is  at 
the  origin  is  not  changed  by  rotating  the  axes  through  any 
angle.  Actually  carry  through  the  transformation. 

The  General  Case.     Let  it  be  required  to  pass  from  one 
system  of  axes  to  a  new  one,  in  which  both  the  origin  and 
the  directions  of   the  axes  have  been 
changed.      Let    the    (x,   y)-axes,   with 
origin  at  0,  be  the  given  system  and 
the  (#',  ?/')-axes,  with  origin  at  0',  the 
new  system.     Let   the   coordinates  of 
0',  referred  to  the  (x,  ?/)-axes,  be  (x0)  y0), 
and  let  the  angle  from  the  positive  axis 
of  x  to  the  positive  axis  of  x'  be  y. 

The  transition  from  one  system  to  the  other  can  be  made  in 
two  steps : 

(a)  Transform  first  to  a  system  of  parallel  axes  having  the 
same  direction,  but  with  origin  at  0'.  If  the  new  coordinates 
are  denoted  by  (X,  Y),  then 


(1) 


I  x  =  X  +  x0, 


(2) 


(&)  Now  rotate  the  (X,  F)-axes  through  the  angle  y : 
[  X  =  x'  cos  y  —  y'  sin  y, 
[  Y  =  xf  sin  y  +  y'  cos  y. 

Combining  these  results  we  get,  as  the  final  formulas,  the 
following : 

/3x  f  x  =  x'  cos  y  -  y'  sin  y  +  xo, 

\y  =  xf  sin  y  +  y'  cos  y  +  y0- 

The  formulas  for  (x',  y')  in  terms  of  (x,  y)  are 


f  x'  =     (x  —  x0)  cos  y  +  (y  —  2/0)  sin  y, 


( y'  =  -  (x  -  XQ)  sin  y  +  (y  -  y0)  cos  y. 


224  ANALYTIC   GEOMETRY' 

These  can  be  written  in  the  form : 
.  .  |  x'  =      x  cos  y  -\-  y  sin  y  —  a\,, 

1  y'  =  —  x  sin  y  -\-  y  c°s  y  —  yo> 
where 

/„  «Q  =      a\>  cos  y  +  y0  sin  y, 

.%  =  —  x0  sin  y  +  y0  cos  y. 

It  is  to  be  noted  that,  inasmuch  as  x0,  y0,  and  y  are  constants, 
so  are  x0  and  y0. 

Example.     Identify  the  curve  represented  by  the  equation 

by  transforming  to  new  axes  through  the  point  (2,  1),  the 
angle  from  the  old  axis  of  x  to  the  new  being  45°. 

Here  x0  =  2,  y0  =  1,  and  y  =  45°.  We  might  substitute  these 
values  in  formulas  (3)  and  then  apply  the  formulas  to  the 
given  equation.  It  is,  however,  more  feasible  in  general  to 
make  the  transformation  in  the  two  steps  (1)  and  (2). 

Formulas  (1)  are,  in  this  case, 

x  =  X  +  2,  y=Y+l. 

Hence  (6)  becomes 

Since  y  =  45°,  formulas  (2)  are  : 

V2  V2 

Then  (7)  becomes 


or,  on  simplification, 


Consequently,    equation    (6)    repre- 
FIG.  7  sents   a   hyperbola  with   its  center  at 


TRANSFORMATION   OF  COORDINATES          225 

the  point  (2,  1)  and  with  its  transverse  axis  inclined  at  an 
angle  of  45°  to  the  axis  of  x. 

In  deducing  formulas  (3),  we  first  shifted  the  origin  and 
then  rotated  the  axes.  We  might  equally  well  have  proceeded 
in  the  opposite  order : 

(a)  Rotate  the  (x,  y)-axes  through  the  angle  y  into  the  new 
axes  of  x  and  y : 

x  =      x  cos  y  +  y  sin  y, 

y  —  ~  x  sin  y  +  y cos  y- 

The  coordinates  of   0',  referred  to 
the  new  axes,  are  obtained  by  setting 


x  =  XQ  and  y  —  y0  in  (8) ;  they  are,  then,  FIG.  8 

the  x0  and  y0  given  by  formulas  (5). 

(6)  Transform  from  the  (a;,  y)-axes  to  the  parallel  axes  of 
a/  and  y',  with  origin  at  0'.  Since  the  coordinates  of  0',  re- 
ferred to  the  (x,  y)-axes,  are  (x0,  y0~),  the  equations  of  this  trans- 
formation are 

f  x'  =  z  —  iT0, 

Eliminating  x  and  ^from  (8)  and  (9),  we  obtain,  as  the  final 
formulas 

x'  =      x  cos  y  +  y  sin  y  —  x0, 

y'  =  —  x  Sin  y  +  ?/  COS  y  —  ?/0. 

But  these  are  precisely  the  formulas  (4)  which  we  had  before. 

EXERCISES 

Obtain  the  equations  of  transformation  in  each  of  the  follow- 
ing cases.  First  find  the  formulas  for  x  and  y  in  terms  of  y! 
and  y',  and  then  solve  for  x'  and  y'  in  terms  of  x  and  y. 

V  0% 2/o)  =  (1>  1) 5  y=45°.         2.  (#0,  ?/0)  =  (  —  2,  1);      y  =  30°. 
'  3.  (x0, y0)  =  (0,3);  y=-60.°    4.  (z0, y0)  =  ( - 5,  -3);  y=120°. 

Draw  a  figure  and  deduce  from  it  directly  the  formulas  of 
transformation  for  each  of  the  following  values  of  y,  the  new 


226  ANALYTIC   GEOMETRY 

origin  in  each'  case  being  at  an  arbitrary  point  (xw  y0).  Check 
the  results  by  the  use  of  formulas  (3)  and  (4)  of  the  text. 

^.    y=90°.  6.     y  =  180°.  7.     y  =  -90°. 

8.  Find  the  coordinates  of  the  points  (0,  0),  (1,  2),  (—3,  4), 
(—2,  —  5),  referred  to  new  axes  passing  through  the  point  (2, 1), 
the  angle  from  the  old  axis  of  x  to  the  new  being  45°. 

9.  The  same,  if  the  new  origin  is  at  the  point  (—3,  4)  and 
the  angle  from  the  old  axis  of  x  to  the  new  is  —  30°. 

Identify  the  curve  represented* by  each  of  the  following 
equations  by  transforming  to  parallel  axes  at  the  point  (x0,  y0) 
specified,  and  then  rotating  the  new  axes  through  the  given 
angle  y.  Draw  a  figure  in  each  case. 

Equation                                   (x0,  y0)  y 

\p.   5a;2-6xy  +  5?/2-4a;-47/-4  =  0,       (1,1),  45°. 

11.  a^-4a;y  +  ^  +  10a!-2y  +  7=  0,          (1,3),  -45°. 

12.  a!2-10ajy  +  y2  +  46aj  +  10y-47  =  0,     (2,5),  135°. 

13.  Find  the  curve  represented  by  the  equation 

66 a£  -  24 xy  +  59 y2  +  108aj  +  94  y  +  76  =  0 
by  introducing  parallel  axes  at  the  point  (—  1,  —  1)  and  then 
rotating  these  axes  through  the  acute  angle  whose  tangent  is  ^. 
Draw  a  graph. 

14.  The  equation 

7x*  -  ISxy  —  17 y2  -  28 a;  +  36y  +  8  =  0 

represents  a  conic  whose  center  is  at  the  point  (2,  0),  and  one 
of  whose  axes  has  the  slope  —  ^.  Identify  the  conic,  and 
draw  a  rough  graph  of  it. 

4.  Determination  of  the  Transformation  from  the  Equations 
of  the  New  Axes.  Consider  the  general  transformation  given 
by  formulas  (4)  of  the  preceding  paragraph.  If,  in  these  for- 
mulas, we  set  x'=  0  and  y'  =  0,  we  obtain 

x  cos  y  +  y  sin  y  -  >r0  =  0, 
—  x  sin  y  +  y  cos  y  —  y0  =  0. 


TRANSFORMATION    OF   COORDINATES  227 

These  are  the  equations  of  the  new  axes,  referred  to  the  old 
system  ;  the  first  is  the  equation  of  the  axis  of  y',  the  second, 
that  of  the  axis  of  x'. 

Conversely,  if  we  set  the  expressions  on  the  left-hand  sides 
of  equations  (1)  equal  to  x'  and  y'  respectively,  we  obtain  the 
equations  of  the  transformation. 

Problem.  Let  it  be  required  to  find  the  equations  of  a  trans- 
formation which  introduces  the  two  perpendicular  lines, 

x  + 
2x-    y-3  =  0, 

as  coordinate  axes. 

The  natural  procedure,  in  order  to  obtain  the  required  equa- 
tions, would  be  to  set  the  left-hand  sides  of  equations  (2)  equal 
to  x'  and  y'.  But,  in  order  to  obtain  the  correct  result  in  this 
way.  we  must  first  put  the  left-hand  sides  of  equations  (2)  into 
the  form  of  those  of  equations  (1). 

These  latter  are  of  the  form 

ax  +  by  -  »0, 
-  bx  +  ay  -  y0, 
where 
(4)  a2  4-  62  =  1. 

The  left-hand  sides  of  equations  (2)  will  be  of  the  form 
(3)  if  we  multiply  the  second  of  the  equations  through 
by-1: 

-2x+ 

To  bring  about  the  fulfillment  of  condition  (4),  we  multiply 
each  of  these  equations  through  by  a  constant  p=£0: 


-2Px  + 

Thereby  we  have  not  changed  the  lines  which  the  original 
equations  (2)  represent. 


228  ANALYTIC   GEOMETRY 

The  value  of  p  is  to  be  determined  so  that  condition  (4)  is  sat- 
isfied by  the  left-hand  sides.  of  equations  (2  a),  that  is,  so  that 


or 


Hence  p  =  ±  I/ V5. 

We  choose  p  =  I/ V5.     If  this  value  is  substituted  for  p  in 
equations  (2  a),  these  equations  will  be  precisely  of  the  form  (1). 
Hence  the  equations  of  a  transformation 
v          lx       introducing  the  lines  (2)  as  the  axes  are 


(5) 


The  first  of  the  lines  (2)  is  the  axis  of  y' ; 
the  second,  the  axis  of  x'. 

The  new  origin  is  at  the  point  (1,  —  1);  for,  this  is  the  point 
of  intersection  of  the  lines  (2).  The  old  origin  (x,  y)  =  (0,  0) 

(1        3  \ 
— ,  — —  ),  referred  to 
V5    V5/ 

the  new  axes,  and  must  lie,  then,  in  the  first  quadrant  formed 
by  these  axes.  Consequently,  the  new  axes  must  be  directed 
as  shown  in  the  figure. 

The  slope  of  the  axis  of  #',  the  second  of  the  lines  (2),  is  2 
and  so  its  slope  angle  is  63°  26',  or  243°  26'.  It  is  clear  from 
the  figure  that  it  is  the  first  of  these  angles  which  is  the  angle  y. 

We  obtain  a  second  transformation,  for  which  the  lines  (2) 
are  the  new  axes,  by  taking  the  value  —  1/V5  for  p.  For  this 
transformation  the  directions  of  both  axes  are  opposite  to  those 
for  (5),  and  y  has  the  value  243°  26'. 

For  both  transformations  the  first  of  the  lines  (2)  is  the  axis 
of  y' ;  the  second,  the  axis  of  x'.  By  reversing  the  roles  of  the 
lines,  two  more  transformations  can  be  obtained.  Thus,  there 
are  in  all  four  transformations  introducing  a  given  pair  of 
mutually  perpendicular  lines  as  coordinate  axes. 


TRANSFORMATION   OF   COORDINATES  229 


EXERCISES 

In  each  of  the  following  exercises  find  the  equations  of  a 
transformation  which  introduces  the  given  perpendicular  lines 
as  new  axes.  Find  the  position  of  the  new  origin  and  the  value 
of  y  ;  draw  an  accurate  figure,  and  indicate  the  directions  of 
the  new  axes. 

Axis  of  y'  Axis  ofx' 


2.  x  +  y  —  3  =  0;  x  —  y  —  1  =  0. 

3.  2x-3y  =  Q.  3x  +  2y  +  5  =  Q. 

4.  5z-2y=0;  2x+5y  =  Q. 

5.  x--2  =  0;  y  +  3  =  0. 

6.  y-8  =  0;  z-5=0. 

7.  If  the  lines  of  Ex.  1  are  introduced  as  axes,  what  does  the 
equation 

(4  a?  -  3y  +  2)2  =  3x  +  4y  -  11 

become  ?     What  curve  does  it  represent  ?     Draw  a  rough  graph. 

By  a  suitable  transformation  of  axes  determine  the  nature 
and  position  of  each  of  the  following  curves.  In  each  case 
draw  a  figure  showing  accurately  the  new  axes,  properly 
directed  ;  then  sketch  the  curve. 

8.     2z-32  +  6z  +  4    +  10  =  0. 
9. 
10. 

11.  (x  +  y  —  I)3  —  5(x  —  y  —  1)=0. 

12.  Find  the  equations  of  all   four  transformations  which 
introduce  the  lines 


5»-12y  +  7  =  0,         12a?  +  5y  —  17  =  0 

as  axes  of  coordinates.     Draw  the  four  corresponding  figures, 
and  find  the  four  values  of  y. 


230  ANALYTIC   GEOMETRY 

13.  Obtain  the  equations  required  in  Ex.  1  by  finding  the 
coordinates  of  the  new  origin  and  a  value  for  y  and  then  ap- 
plying formulas  (3)  of  §  3. 

5.   Reversal  of  One  Axis.     There  is  one  more  case  which 
deserves  mention.     Suppose  that  the  sense  of  one  axis  is  re- 
versed, while  the  other  axis  remains  un- 
changed.     Let  the  axis  which  is  reversed 
be  the  axis  of  x  (Fig.  10).     Then,  evidently,    ' 

X  — —  ?  I         """*""  j 

' 


O        M 
FIG.  10  (1) 

( y  • —  y  i  (J       J~ 

If  the  sense  of  the  axis  of  y  had  been  reversed,  the  axis  of 
x  remaining  unchanged,  we  should  have  had  : 

/ON  I  fl!  •—  «B  j  j  X   =  X, 

Consider,  for  example,  the  equation  of  a  parabola  in  the 
normal  form, 

y-  =  2  mx. 

If  the  sense  of  the  axis  of  x  is  reversed,  the  equation  becomes 

y'2  =  -  2  mx'. 

We  could  use  this  result  to  interpret  the  equation, 
y '  ^^  —  Zt  TYioC) 

if  we  knew  the  parabola  only  in  its  normal  form.  Taking  the 
axis  of  x'  opposite  to  the  axis  of  x,  and  starting  with  the 
known  parabola 

y'-  =  2  mx', 

we  see  that  the  transformed  equation, 
y1  =  —  2  mx, 

represents  a  parabola  on  the  negative  axis  of  x, 

its  vertex  being  at  the  origin.  FIG.  11 


TRANSFORMATION   OF   COORDINATES  231 

EXERCISES 

1.  Assuming  that  the  equation  of  the  parabola  in  the  form  x 

x2  =  2  my 
is  known,  interpret  the  equation 

xz  =  —  2  my 
by  the  method  of  the  text. 

2.  Plot  the  so-called  semi-cubical  parabola 

f-  =  y?. 
From  the  graph  determine  the  curve  denned  by  the  equation 


EXERCISES  ON  CHAPTER  XI 
CHANGE  OF  ORIGIN 

In  each  of  the  following  exercises  prove,  by  making  a  suit- 
able transformation  to  parallel  axes,  that  the  given  equation 
represents  two  straight  lines.  Find  the  equations  of  the  lines, 
referred  to  the  original  axes. 


2    4z2  —  924-8z  +  18    —  5  =  0. 
3.   4z> 


What  does  each  of  the  following  equations  represent  ? 
4.   z2-f  2y*-Wx  +  12y  +  43  =  0. 

-4ns.    The  point  (5,  —  3). 
5. 


6.    By  completing  the  cube  for  the  terms  in  x  in  the  equa- 
tion 


and  by  making  the  transformation  to  parallel  axes  which  is 
suggested  by  the  result,  determine  the  curve  defined  by  the 
equation.  Draw  the  curve  roughly. 


232  ANALYTIC   GEOMETRY 

By  the  method  of  Ex.  6,  identify  and  plot  the  locus  of  each 
of  the  following  equations. 

7.  6y  =  x* 

8.  x  =  2y* 

9.  x3  —  3a2 


10.   Determine  the  position  of  the  point  (x0,  y0)  such  that,  if 
the  straight  lines 


are  referred  to  parallel  axes  at  (x0,  ?/0),  their  equations  will 
contain  no  constant  terms.  What  will  these  equations  be  ? 

11.  Show  that,  if  the  two  intersecting  straight  lines 

A&  +  £#+<?!  =  0,        A2x  +  Biy+C2  =  Q 

are  referred  to  parallel  axes  at  their  point  of  intersection,  the 
equations  of  the  lines  become 

A,?!  +  Biy'  =  0,        A&1  +  B<$'  =  0. 

12.  Determine  the  position  of  the  point  (x0,  t/0)  such  that,  if 
the  curve 

xy  -2x  —  y  —  2  =  0 

is  referred  to  parallel  axes  at  (x0,  y0),  it§  equation  will  contain 
no  linear  terms  in  x  and  y.  Identify  and  plot  the  curve. 

13.  Identify  and  plot  roughly  the  locus  of  the  equation 

6xy  +  7x  —  5y  +  3  =  0. 

14.  Determine  the  point  (x0,  ?/0)  such  that,  if  the  curve 

3xz  -  7xy  -  6y2  -  19z  +  2y  +  20  =  0 

is  referred  to  parallel  axes  at  (£0,  ?/0),  its  equation  will  contain 
no  linear  terms  in  x  and  y.  Show  that  the  equation  will  also 
contain  no  constant  term  and  hence  that  it  will  represent  two 
straight  lines.  Find  the  equations  of  these  lines  with  respect 
to  the  original  axes. 


TRANSFORMATION   OF   COORDINATES  233 

ROTATION  OF  AXES 

15.  Given  the  two  perpendicular  lines  through  the  origin 
of  slopes  ^  and  —  2.     Find  the  equations  of  a  transformation 
introducing  these  lines  as  axes. 

16.  The  equation  2xz  -\-3xy  —  2y2  =  0  represents  two  per- 
pendicular lines  through  the  origin.     Show  that  it  may  be 
transformed  into  the  equation  x'y'  =  0  by  a  suitable  rotation 
of  axes. 

Identify  and  plot  roughly  the  curve  defined  by  each  of  the 
following  equations.     Cf.  Ex.  16. 

17.  2x2  +  3xy  —  2y*  =  Q.  18.    12x2  -  7  xy  —  12y2=25. 


19.  Determine  the  equations  of  a  rotation  of  axes  where- 
by   the    axis    of    y    comes    into    coincidence    with    the    line 
4z  +  3?/  =  0. 

20.  Identify  and   plot   roughly  the   curve  denned  by  the 
equation  (4  x  +  3?/)2  =  125  x.     Cf  .  Ex.  19. 

The  same  for  each  of  the  following  equations. 

21.  (x  +  t/)2  =  4V2y. 

22.  4aj2  +  4o*/  -f-  y2  = 
23. 


24.    The  line  2x  —  y  —  0  is  an  axis  of  the  conic 


By  a  suitable  rotation  of  axes  determine  the  nature  and  posi- 
tion of  the  conic. 

Show  that,  by  a  suitable  rotation  of  axes,  each  of  the  follow- 
ing equations  becomes  linear  in  y  and  hence  capable  of  solu- 
tion for  y,  without  radicals. 

25.  z2-y2  +  2a;-3  =  0. 

26.  4ic2—  4as    +    2  +  3x  —     =  2. 

27.  2x'2 


234  ANALYTIC   GEOMETRY 

GENERAL  TRANSFORMATION  OF  AXES 

28.  Prove  that  the  straight  lines 

3x  -4y  — 2  =  0,         x  +  2y  =  ±, 

when  referred  to  suitable  axes,  will  have  equations  the  first 
of  which  is  x'  =  0,  while  the  second  contains  no  constant  term. 

29.  Find  the  equations  of  the  circle 

z2_|_y2_|_6a;-Sy-t-6  =  0 
and  the  line  5x  +  12y  —  13  =  0, 

when  they  are  referred  to  axes  through  the  center  of  the  circle 
parallel  and  perpendicular,  respectively,  to  the  line. 

30.  Find  the  equations  of  the  two  circles 

a2  +  y2  +  4x  —  6y  —  4  =  0, 
x~  +  y2  -  6 x  +  4y  —  4  =  0, 

when  they  are  referred  to  the  point  midway  between  their 
points  of  intersection  as  origin  and  the  line  joining  the  points 
of  intersection  as  axis  of  x'. 

31.  What  will  the  equations  of  the  circles  of  Ex.  30  become, 
if  they  are  referred  to  the  mid-point  between  their  centers  as 
origin  and  the  line  of  the  centers  as  axis  of  y'  ? 

32.  A  transformation  consists  of  a  change  of  origin  to  the 
point  (x0,  y0),  of  a  rotation  of  the  new  axes  through  the  angle 
y,  and  of  a  reversal  of  the  sense  of  the  axis  of  x  thus  obtained. 
Show  that  the  equations  of  the  transformations  are. 

x  =  —  x'  cos  y  —  y'  sin  y  +  #o> 
=  —  x'  sin  y  +  y'  cos  y  +  y0. 


CHAPTER   XII 


THE  GENERAL  EQUATION  OF  THE   SECOND  DEGREE 

1.  Change  of  Origin  of  Coordinates.  The  aim  of  this  chapter 
is  twofold :  To  determine  what  curves  are  represented  by 
equations  of  the  second  degree  in  x  and  y ;  and  to  develop 
methods  by  means  of  which  the  curve  represented  by  any 
particular  equation  may  be  easily  identified  and  its  size  and 
position  accurately  described.  The  methods  used  consist 
primarily  in  transformations  of  coordinates.  We  begin,  then, 
by  investigating  what  can  be  accomplished  by  a  change  of 
origin,  i.e.  a  transformation  to  parallel  axes. 

Example  1.  Let  it  be  required  to  identify  and  to  describe 
accurately  the  curve  represented  by  the  equation 

Completing  the  square  of  the  terms  in  x  and  then  of  the 
terms  in  y,  according  to  the  method  of  Ch.  XI,  §  1,  we  obtain : 

5  (a-  _  2)2  -  4  (y  +  3)2  =  -  20. 
On  setting 

x'  =  x  -  2,  yf  =  y  +  3, 

that  is,  on  changing  the  origin  of  coordi- 
nates to  the  point  (2,  —  3),  this  equation 
becomes 

Sa/2  — 4y'2  =  —  20, 

QJ* . y_ — i 

4         5  FIG.  1 

Consequently,  equation  (1)  represents  a  hyperbola  with  its 
center  at  the  point  (2,  —  3)  and  with  its  transverse  axis  par- 

235 


236 


ANALYTIC   GEOMETRY 


allel  to  the  axis  of  y.  The  coordinates  of  the  foci,  referred 
to  the  new  axes,  are  (0,  ±  3)  ;  consequently,  when  referred 
to  the  original  axes,  they  are  (2,  0)  and  (2,  —  6).  The  equa- 
tions of  the  asymptotes,  with  respect  to  the  new  axes,  are 


hence  they  are,  with  respect  to  the  original  axes, 


The  semi-axis  major  is  V5,  and  the  semi-axis  minor,  2  ;  the 
eccentricity  has  the  value  f  VB.  . 

Example  2.     Consider  the  equation 
(2)  3xy-6'x  +  3y-10  =  0. 

We  rewrite  this  equation,  first,  in  the  form 

3(xy-2x  +  y        )=10, 
and  then  as  3  [x  (y  —  2)  +  (y        )]  =  10. 

If  —  2  is  added  to  the  y  in  the  second  parenthesis  and,  in 
equalization,  3-1  -(—2)  or  —6  is  added  to  the  right-hand 
side,  this  equation  becomes 


We  now  change  the  origin  to  the  point 
f/     (-  1,  2)  by  setting 


=  x 


-x       The  equation  thus  becomes 


Fio.  2 


Accordingly,  (2)  represents  a  rectangular  hyperbola  with  the 
lines  x  +  1  =  0  and  y  —  2  =  0  as  asymptotes. 

Example  3.     The  equation. 


(3)  z2  +  2o;-2?/-l  =  0, 

can,  according  to  the  method  of  Ch.  XI,  §  1, 
be  put  into  the  form 


o' 
FIG.  3 


EQUATION   OF   THE   SECOND   DEGREE          237 

and   hence   represents  a  parabola  with  vertex   at   the   point 
(—  1,  —  1)  and  with  axis  parallel  to  the  axis  of  y. 

Example  4.  Consider,  now,  an  equation  in  which  all  three 
quadratic  terms  are  present : 

(4)  6x2-xy-2y*  +  4;X  +  9y-W  =  Q. 

In  this  case,  completing  the  squares  of  the  terms  6 a;2 +  4  a; 
and  of  the  terms  —  2y2+9y  does  not  help.  Let  us  make  an 
arbitrary  change  of  origin,  setting 

(5)  x  =  x'  +  x0,         y  =  y'+y0, 

and  aim  to  determine  the  new  origin,  (x0,  y0),  so  that  in  the  re- 
sulting equation  the  linear  terms  in  x'  and  y'  do  not  appear. 

Setting  in  (4)  the  values  of  x  and  y  as  given  by  (5)  and 
collecting  terms,  we  have 

(6)  6x'*-x'y'  -2ylz 

+  (12  a-0  -  y0  +  4)z'  +(-  x0  -  4y0  +  9)</'  +  F'  =  0, 
where 

(7)  F'  =  6 z02  -  a^0  -  2?/o2  +  4a>0  +  9y0-  10. 

If  the  terms  in  x'  and  y'  are  to  drop  from  this  equation,  x0  and 
y0  must  be  so  chosen  that 

12a?0-    ?/0  +  4  =  0, 

-z0-42/0  +  9  =  0. 

Solving  (8)  simultaneously  for  x0,  y0,  we  have 

ZQ  =  -  |,         y0  =  -V1 . 

The  value  of  F',  for  these  values  of  XQ  and  yQ,  is  0.  Conse- 
quently, we  have  shown  that  equation  (4),  when  referred  to  a 
new  origin  at  (—  -f,  2|),  becomes 

6xn--x'y'-2y'2  =  Q. 

The  left-hand  side  of  this  equation  can  be  factored : 
(9)  (3x'-2y')(2x'+y')=Q. 

Thus  (4)  represents  two  straight  lines  through  the  point 
(-  |,  2f)  with  slopes  f  and  -  2. 


238  ANALYTIC   GEOMETRY 

If  in  (9)  we  set 


i.e.  if  we  transform  back  to  the  original  axes,  we  obtain,  finally, 


This  equation  is  seen  to  be  precisely  equation  (4),  with  the 
left-hand  side  factored  into  two  linear  factors.  The  two 
straight  lines  represented  by  (4)  have,  then,  the  equations 

3x-2y  +  5  =  Q,        2x  +  y  —  2  =  Q. 

It  should  be  noted  that  the  constant  term  (7)  of  equation  (6) 
is  the  value  of  the  left-hand  side  of  (4)  for  x  =  xw  y  =  y0.  In 
this  example,  this  constant  term  took  on  the  value  zero  when 
x0,  y0  were  chosen  so  that  the  coefficients  of  the  linear  terms  in 
x'  and  y'  vanished.  This  does  not,  however,  occur  in  general, 
as  we  shall  see  in  the  next  paragraph. 

EXERCISES 

In  each  of  the  following  exercises  identify  and  plot  roughly 
the  curve  represented  by  the  given  equation.  If  the  curve  is 
a  conic  section,  find,  when  they  exist,  the  coordinates  of  the 
center  and  the  foci,  the  equations  of  the  directrices  and  the 
asymptotes,  and  the  value  of  the  eccentricity. 


2.  18x*  +  12y*-12x  +  12y-  19  =  0. 

3.  4  xz  -f  3  y*  +  16  x  —  6y  +  31  =  0.  Ans.     No  locus. 
4. 

5. 

6.  7x"-  —  5y-  +  2x  —  4y  —  1  =  0. 

7.  y1-  8x  +  6  y  +  49  =  0. 

8.  3  x*-  —  6x  -5?/-f3  =  0. 

9.  2^+4z  +  3?/-8  =  0. 

10.  xy+  2x-3y  -11  =  0. 

11.  5xy  —  5x  +  y  -f  1  =0. 


EQUATION   OF   THE   SECOND   DEGREE          239 

12.  3xy  +  x  —  18y  —  6  =  0. 

13.  2x'i+5xy-3y*  -\-  3x  +  I6y-  5  =  0. 

14.  xn-  +±zy  +  3if  —  2x  —  2y  =  0. 

15.  3*'  —  xy  +  5y2  —  6x  +  y-f  3  =  0.     Ans.  The  point  (1,  0). 

16.  Prove  that  every  equation  of  the  form 

y  =  Ax*  +  Bx  +  (7,  ^1  =£  0, 

or  of  the  form 


represents  a  parabola  with  its  axis  parallel  to  an  axis  of  coor- 
dinates. 

17.  Show  that  every  equation  of  the  form 

bxy  +  dx  +  ey  +/=  0,  b  =£  0, 

represents  either  a  rectangular  hyperbola  with  its  asymptotes 
parallel  to  the  axes,  or  two  perpendicular  straight  lines  parallel 
to  the  axes.  Prove  that  the  latter  case  occurs  if  and  only  if 
bf=  de. 

18.  Given  the  equation 

ax2  +  cy2  +  dx  +  ey  +/=0, 

where  neither  a  nor  c  is  0  :  ac  ^  0. 

(a)  If  ac  >  0,  prove  that  the  equation  represents  an  ellipse, 
or  a  point,  or  that  it  has  110  locus. 

(6)  If  ac  <  0,  show  that  the  equation  represents  a  hyperbola, 
or  a  pair  of  intersecting  straight  lines. 

2.   Rotation  of  Axes.     Example  1.     Let  it  be  required  to 

identify  the  curve  defined  by  the  equation 

(1)  5x*-Gxy  +  5y*-8  =  Q. 

We  transform  (1)  by  a  rotation  of  the  (x,  y)-axes  through 
an  arbitrary  angle  y  into  the  (x',  y')-axes.  For  x  and-y  in  (1) 
we  set,  then,  according  to  Ch.  XI,  §  2, 

2.  x  =  x'  cosy  -y'  sin  y, 

y  =  x'  sin  y  +  y'  cos  y, 


240 


ANALYTIC   GEOMETRY 


and  obtain,  after  collecting  terms  and  simplifying, 

(5  —  6  cos  y  siny)o312  —  6x'y'(GOSz  y  —  sin2  y)  +-(5  +  6  sin  y  cos  y)y'2 

-  8  ='o, 
or,  on  replacing  the  trigonometric  functions  of  y  by  functions 

Of2y, 

(3)  (5-3  sin2y)o;'2  -  Qx'y'  cos 2y  +(5  +  3 sin2y)y'2 -8  =  0. 
We  now  choose  y  so  that  the  coefficient  of  x'y'  will  become  0 : 
cos  2  y  =  0. 

Values  of  2  y  satisfying  this  equation  are  90°,  270°,  450°,  630° ; 
the  corresponding  values  of  y  are  45°,  135°,  225°,  315°.  We 
choose,  arbitrarily,  the  smallest  of  these  values,  namely,  y=45°. 
Equation  (3)  thus  becomes 


or 

T  +  T=:  ' 

Consequently,  equation  (1)  represents  an  ellipse  with  its 
center  at  the  origin  and  with  the  transverse  axis  inclined  at  an 
angle  of  45°  to  the  axis  of  x. 

Example  2.     Consider  the  equation 

We  proceed,  as  in  §  1,  Example  4,  transforming  (5)  to  arbi- 
trary parallel  axes,  x,  y,  and  then  choosing  the  new  origin, 
(x0,  y0),  so  that  in  the  equation  resulting 
from  (5)  the  linear  terms  in  x  and  y  drop 
out.  We  find  that  the  new  origin  must  be 
at  the  point  (1,  1),  and  that  the  resulting 
equation  then  becomes 

F'1G  4  where  the  constant  term,  —  8,  is  found  as 

the  value  of  the  left-hand  side  of  (1)  for 
x  =  1,  y  =  1 ;  cf .  end  of  §  1. 

Now  (6)  is  the  same  equation  in   x,   y  as  (1)    is   in  #,   y. 


EQUATION   OF   THE    SECOND   DEGREE  241 

Hence,  it  follows  that  (5)  represents  an  ellipse  with  its  center 
at  the  point  (1,  1)  and  with  the  transverse  axis  inclined  at  an 
angle  of  45°  to  the  axis  of  x. 

The  procedure,  then,  for  any  equation  similar  in  form  to  (5) 
consists  first  in  transforming  to  parallel  axes  so  that  the  linear 
terms  in  x  and  y  drop  out,  and  then  in  rotating  the  new  axes 
so  that  the  quadratic  term  in  #,  y  drops  out.  We  shall  show 
later  that  this  procedure  is  always  valid  except  in  one  case. 

EXERCISES 

Identify  the  curves  represented  by  the  following  equations. 
Draw  a  graph  in  each  case,  showing  the  original  and  the  new 
axes  and  the  curve. 


2.  5 

3.  lxz  +  2xy  +  7  ?/2  +  2  =  0. 

4.  6  a2  +  2  ^/3xy  +  7y*  —  16  =  0. 

5.  2  x2  +  4  V3xy-  2  y2  —  16  =  0. 

6.  3a2  —  2xy  +  3y2  —  4x  —  4#  =  0. 

7.  x*  +  6xy  +  y1  -  Wx  —  14  y  +  14  =  0. 

8.  4a2  +  16a*/  +  4?/2  —  4z  —  8y  +  13  =  0. 

9.  Show  that,  if  6  =£  ±  2  a,  the  equation 

ax*  +  bxy  +  ay2  +  /  =  0,  6/=jt  0, 

represents  an  ellipse,  or  a  hyperbola,  with  its  center  at  the 
origin  and  with  its  axes  bisecting  the  angles  between  the 
coordinate  axes. 

3.  Continuation.  General  Case.  We  propose  to  develop  and 
simplify  the  method  of  §  2,  Example  1,  for  the  removal  of  the 
term  in  xy.  Take  the  equation 

(1)  Ax2  +  Bxy  +Cy*  +  F'  =  Q,  B  =£  0, 

and  rotate  the  axes  through  the  arbitrary  angle  y,  by  means  of 
formulas  (2),  §  2.  The  resulting  equation  can  be  written  as 


242  ANALYTIC    GEOMETRY 

(2)  ax'*  +  bx'y'  +  cy'2  +  F'  =  0, 
where 

a  =  A  cos2  y  +  B  sin  y  cos  y  +  Cy  sin2  y, 

(3)  6  =  -(^4-  <7)sin2y  +  .Bcos2y, 

c  =  A  sin2  y  —  B  sin  y  cos  y  +  C  cos2  y. 

Since  y  is  to  be  chosen  so  that  b.  =  0, 

(4)  -(A-  <7)sin2y  +  .Bcos2y  =  0, 

(5)  cot2y  =  ^^'. 

B 

Of  the  values  of  2y  which  satisfy  this  equation,  we  choose 
arbitrarily  that  one  which  lies  between  0°  and  180°.  Then  y 
is  a  positive  acute  angle. 

If  the  axes  are  rotated  through  this  angle  y,  (2)  becomes 

(6)  ax'2  +  cy'*  +  F'  =  Q. 

The  values  of  a  and  c  are  still  to  be  determined.  There  is 
a  simpler  way  of  doing  this  than  substituting  the  value  found 
for  y  in  the  formulas  for  a  and  c,  as  given  by  the  first  and  last 
equations  of  (3).  First,  add  these  two  equations  ;  the  result  is 

(7)  a  +  c  =  A+C. 

Thus  we  have  one  very  simple  equation  for  the  two  unknown 
quantities  a  and  c.  , 

Next,  subtract  the  second  of  the  two  equations  from  the 
first: 

(8)  a  -  c  =  (A  -  C)  cos  2y  +  B  sin  2  y. 

Square  both  sides  of  (8)  and  both  sides  of  the  second  equation 

Of  (3)  : 

b  =  -  (A-  C)  sin2y+  Bcos2y, 

and  add  the  equations  thus  obtained ;  the 'final  result  is 

(9)  (a  -  c)2  +  &2  =  (A  -  C)2  +  B2. 

But  y  was  chosen  so  that  6  =  0.     Consequently,  (9)  becomes 


EQUATION   OF  THE   SECOND   DEGREE          243 

or 

(10)  a-c  =  ±  V(4  -  O)2  +  &. 

Thus  we  have  a  second  simple  equation  for  the  two  unknowns 
a  and  c. 

From  equations  (7)  and  (10)  the  values  of  a  and  c  are  easily 
found  in  terms  of  the  known  coefficients,  A,  B,  and  (7,  of  (1). 
There  are,  however,  two  values  of  each,  due  to  the  double 
sign  before  the  radical  in  (10).  Which  values  should  we 
take? 

If  in  (8)  we  substitute  for  A  —  C  its  value  as  given  by  (4), 
we  obtain 

a  -  c  =  B  cos2  2  y  +  B  sin  2y, 

sin2y 
or 

B 


a  —  c  = 


sin  2- 


But  2  y  lies,  by  choice,  between  0°  and  180°  and,  consequently, 
sin  2  y  is  positive.  It  follows  that  a  —  c  must  have  the  same 
sign  as  B. 

Accordingly,  if  we  rewrite  (10)  as 


the  plus  sign  must  be  chosen.     Hence,  always, 
(11)  a-c  = 


From  equations  (7)  and  (11)  unique  values  for  a  and  c  can 
now  be  found. 

Example,.     Consider  the  equation 

(12)  7x*-Sxy  +  yz  +  Ux-8y-2  =  Q. 

By  shifting  the  origin  properly,  to  the  point  (—  1,  0),  (12) 
becomes 

(13)  7  a/2  -  8ajy  +  y'2  -  9  =  0. 


244 


ANALYTIC   GEOMETRY 


Next,  rotate  the  new  axes  through  the  positive  acute  angle 
given  by  formula  (5),  which  in  this  case  is 

7-1         3 

COt  2  y  =  . —  =  —  -, 

so  that  y  has  the  value  63°  26'.     Thus  (13)  becomes 
(14)  ax"2  +  cy"*  -9  =  0. 

The  values  of  a  and  c  are  determined  from  equations  (7) 
and  (11),  which  are,  here, 

a  +  c  =  8, 


a  —  c  =  — 


Then  the  values  for   a  and   c  are :    a  =  —  1,   c  =  9.     Conse- 
quently, (14)  becomes 


FIG.  5 


*      y_ i 

~9~"T~ 

Equation  (15)  represents  a  hyperbola 
with  its  transverse  axis  along  the  axis  of 
y".  Hence  (12)  represents  a  hyperbola 
with  its  center  at  (—1,  0)  and  with  its 

transverse  axis  inclined  at  an  angle  of  63°  26'  +  90°  =  153°  26' 

with  the  axis  of  x. 

TJie  Expression  B2  —  4 AC.  If  from  (9)  we  subtract  the 
square  of  (7),  we  obtain 

or,  since  we  chose  y  so  that  6  =  0, 

(17)  -4ac  =  B2 

The  Case  B*  -  4  AC  >  0.  If  B2  -  4  AC  is  positive,  ac  is,  by 
(17),  negative ;  hence  a  and  c  have  opposite  signs.  Thus,  if 
F'  =£  0,  (6)  represents  a  hyperbola.  If  F'  =  0,  (6)  becomes 

(18)  ax'2  +  cy'2  =  0. 


EQUATION   OP  THE    SECOND   DEGREE          245 

Since  a  and  c  have  opposite  signs,  the  left-hand  side  of  (18) 
can  be  written  as  the  difference  of  two  squares  and  then  fac- 
tored. Therefore,  (18)  represents  two  straight  lines  (Ch.  IX, 
§  4)  which  intersect  at  the  origin. 

These  results  for  (6)  are  true  for  the  original  equation  (1). 
They  hold  not  only  if  B=£Q,  —  the  case  which  we  have  been 
treating,  —  but  also  if  .6  =  0.  For,  if  2?  =  0,  (1)  is  itself  in 
the  form  (6),  and  hence  may  be  considered  directly.  We  have, 
then,  the  following  theorem. 

THEOREM  1.      When  .B2  —  4  AC  >  0,  the  equation 
Ax*  +  Bxy  +  Cf-  +  F'  =  0 

represents  a  hyperbola,  if  F'^Q;  ifF'  =  Q,  it  represents  two 
intersecting  straight  lines. 

The  Case  B2  —  4:AC<0.  In  this  case,  according  to  (17), 
ac  is  positive,  and  a  and  c  have  the  same  signs.  Then  if 
F'  3=  0,  (6)  represents  an  ellipse,  or,  in  the  case  that  a,  c,  and  F1 
are  all  of  the  same  sign,  has  no  locus.  If  Fr  —  0,  (6)  reduces 
to  (18).  But  now  the  left-hand  side  of  (18)  can  be  written 
as  the  sum  of  two  squares,  since  a  and  c  have  the  same  sign. 
Hence  it  is  satisfied  only  by  x  =  0,  y  =  0.  It  represents,  then, 
a  single  point,  or,  as  we  may  say,  a  null  ellipse.* 

Not  merely  a  and  c  have  the  same  signs  in  this  case,  but 
also  A  and  C.  For,  if  A  and  C  have  not  the  same  signs,  the 
product  AC  <  0  or  =  0 ;  consequently,  B2  —  4  AC  >  0,  —  a  con- 
tradiction. It  follows,  further,  from  (7),  that  A  and  C  have 
the  same  signs  as  a  and  c. 

We  can  now  characterize  more  fully  the  two  cases  which 
arise  when  F'=£Q.  We  have  seen  that  equation  (6)  has  no 
locus,  if  F'  is  of  the  same  sign  as  a  and  c,  or,  as  we  can  now 
say,  if  F'  is  of  the  same  sign  as  A  and  C,  i.e.  if  AF'  (or 
CF'}  >  0.  On  the  other  hand,  (6)  represents  an  ellipse,  if  F'  is 
opposite  in  sign  to  A  and  C,  i.e.  if  AF'  (or  CF')  <  0. 

We  summarize  our  results  in  the  form  of  a  theorem. 

*  Cf .  null  circle,  Ch.  IV,  §  2. 


246  ANALYTIC   GEOMETRY 

THEOREM  2.      When  B2  —  4:  AC  <  0,  the  equation 
Bxy  +  Cyz  +  F'  =  0, 


if  F'  =£.  0,  represents  ah  ellipse  or  has  no  locus,  according  as  AF' 
(or  OF")  is  negative  or  positive;  if  F'  =  0,  the-  equation  repre- 
sents a  single  point 

The  Case  .B2-  4.4(7=0.     If  &  -4.4(7=0,  there  is  no 
need  of  rotating  the  axes.     Consider,  for  example,  the  equation 

(19)  9o;2 


for  which  B2  —  4  AC  =  36  -  4  •  9  =  0.     This  equation  can  be 
written  in  the  form 


or 

and  hence  represents  two  parallel  lines  of  slope  3. 

EXERCISES 

Identify  the  curves  represented  by  the  following  equations. 
Draw  a  graph  in  each  case,  showing  the  original  and  the  new 
axes  and  the  curve. 


=0. 

2.  3xz  +  12xy  +  8yz  +  6x  +  16y  +  38  =  0. 

3.  73aj2+72xy  +  52y*  +  7<lx-32y  —  47  =  0. 

4.  2x2  +  3xy-2y2-  16z  -  12?/  +  22  =  0. 

5.  *2  —  5xy  +  13y*-3x  +  21y  =  0. 

6.  15  xy-Sy*  +  450^-450  =  0. 

7.  20z2-16a^  +  8y2  +  52z-40?/  +  5  =  0. 

8.  8x*  +  8xy-  7^  +  36^  +  36  =  0. 

9.  7  x2  -  3  xy  +  3y2  +  5x  +  I5y  +  35  =  0.    Ans.   No  locus. 
10. 

11. 

12.  xz  +  3xy  —  yz  +  2x—  Wy  =0. 


EQUATION   OF  THE   SECOND   DEGREE          247 

4.   The  General  Equation,  IP  —  4  AC  =£  0.     We  consider  here 
the  general  equation  of  the  second  degree  : 

(1)  Ax2  +  Bxy  +Cf  +  Dx  +  Ey  +  F  =  Q, 

assuming  that  B2  —  4  AC  =£  0.  From  the  results  of  the  preced- 
ing paragraph,  we  should  expect  that,  in  general,  (1)  represents 
an  ellipse  or  has  no  locus,  if  B2  —  4  AC  <  0,  and  represents 
a  hyperbola,  if  B2  —  4  .4(7  >  0.  Accordingly,  we  shall  call 

(1)  an  equation  of  elliptic  type  or  of  hyperbolic  type,  according  as 
W  —  4  .4(7  is  negative  or  positive.* 

To  remove  the  terms  in  x  and  y  from  (1),  we  set 

x  =  x'  +  ZQ,          y  =  y'  +  y0 
in  (1),  obtaining 

(2)  Ax'2 


where 

(3)  F'  = 

is  the  value  of  the  left-hand  side  of  (1),  formed   for  x  =  050, 

2/  =  2/o- 

Setting  the  coefficients  of  x'  and  y'  in  (2)  equal  to  zero  : 

xx  2Ax0+     By0  +  D  =  0, 


and  solving  these  equations  simultaneously  for  x0  and  y0,  we 
have 

,-  _2CD-BE  2AE-BD 

°~ 


Since  it  has  been  assumed  that  the  denominator,  B2  —  4  4(7, 
of  these  fractions  is  not  0,  it  is  always  possible  to  solve  equa- 
tions (4),  and  the  solution  (5)  is  unique. 

If  the  new  origin  (x0,  y0)  is  taken  at  the  point  (5),  equation 
(2)  becomes 
(6)  Ax'2  +  Bx'y'  +  Cy'2  +  F'  =  0. 

*  If  B2  -  1AC  -  0,  we  shall  say  that  (1)  is  of  parabolic  type.    This 
case  will  be  treated  in  the  next  paragraph. 


248  ANALYTIC   GEOMETRY 

Equation  (6)  is  exactly  the  equation  treated  in  §  3.  There- 
fore the  theorems  of  §  3  are  valid  for  it  and,  consequently,  for 
the  original  equation  (1). 

The  value  of  F'  as  given  by  (3)  can  be  put  in  a  more  con- 
venient form.  Multiply  the  first  of  the  equations  (4)  by  x0, 
the  second  by  y0,  and  add : 

2  Axf  +  2  BXMO  +  2  Cy?  +  Dx,,  +  Ey0  =  0. 
Multiply  this  equation  by  —  ^  and  add  it  to  (3) : 


Finally,  substitute  the  values  of  x0  and  y0  as  given  by  (5). 
The  result  is 

F,  =      4:ACF-  &F  -  AE2  -  CD2  +  BDE 
B*-4:AC 

The  numerator  of  the  fraction  is  known  as  the  discriminant 
of  equation  (1)  and  is  denoted  by  A : 

(7)  A  =  4  ACF—  B*F  —  AN- 

In  terms  of  A,  F'  has  the  value 

(8)  F'  =  -         A 


B*-±AC 

It  is  clear  that  if  F'  =  0,  then  A  =  0,  and  conversely.  In 
stating  the  theorems  of  §  3  for  equation  (1)  above,  we  can, 
therefore,  replace  F'  =£  0  and  F'  =  0  by  A  =£  0  and  A  =  0  respec- 
tively. Furthermore,  in  case  .B2  — 4  AC  is  negative  and  F' 
and  A  are  not  0,  A  has  the  same  sign  as  F'.  In  this  case, 
then,  AF'  (or  CF')  is  positive  or  negative,  according  as  AA 
(or  CA)  is  positive  or  negative. 

We  now  restate,  for  equation  (1),  the  theorems  of  §  3. 

THEOREM  3.     An  equation  (1)  of  hyperbolic  type : 


represents  a  hyperbola,  if  A  =£  0.     If  A  =  0,  it  represents  two 
intersecting  straight  lines. 


EQUATION    OF   THE    SECOND    DEGREE  249 

THEOREM  4.     An  equation  (1)  of  elliptic  type  : 


if  A  =£  0,  represents  an  ellipse  or  has  no  locus,  according  as  A&. 
(or  CA)  is  negative  or  positive  ;  if  A  =  0,  the  equation  represents 
a  single  point. 

If  an  equation  of  the  form  (1)  is  given,  and  IP  —  4  AC  =£  0, 
the  type  of  curve  which  the  equation  represents  can  be  de- 
termined by  finding  the  sign  of  B2  —  4  AC  and  by  ascertaining 
whether  or  not  A  =  0.  Further  investigation  is  necessary 
only  in  case  B2  —  4  AC  <  0  and  A  =£  0  ;  the  sign  of  A  A  (or  CA) 
must  then  be  determined. 

For  example,  the  equation 

a2  —  3xy  +  2yi  +  x  —  5y  +  3  =  0 

represents  a  hyperbola,  inasmuch  as  .B2—  4.4(7=9—  4  •  2=1>0, 
and  A  =  -  15  =£  0. 

To  find  the  position  and  size  of  an  ellipse  or  a  hyperbola 
defined  by  an  equation  of  the  form  (1),  it  is  necessary  to  carry 
through  in  detail  the  work  of  changing  the  origin  and  rotating 
the  axes.  If,  however,  A  =  0,  it  is  sufficient  merely  to  make 
the  proper  change  of  origin.  The  equation  then  takes  on  the 
form  (6),  where  F'  —  0.  In  the  elliptic  case,  it  represents  a 
single  point,  the  new  origin.  In  the  hyperbolic  case,  it  can 
be  factored  into  two  linear  equations,  which  determine  the 
two  lines  typical  of  this  case. 

EXERCISES 

Determine  the  nature  of  the  curve  defined  by  each  of  the 
following  equations.  In  case  the  equation  represents  two 
straight  lines  or  a  single  point,  find  the  equations  of  the  lines, 
or  the  coordinates  of  the  point,  referred  to  the  (x,  ?/)-axes. 

1.  4a?  —  5xy 

2.  3xi 

3.  3x* 

Ans.   The  point  (1,  1). 


250  ANALYTIC    GEOMETRY 


4.  2x*  +  3xy  —  2y*  —  lla;  -  2t/  +  12  =  0. 

Ans.    The  lines  2x  —  y  —  3  =  0,     a  +  2y  —  4  =  0. 

5.  x*  +  xy  +  y-  +  3y  +  4  =  0.  Ans.   No  locus. 

6.  3a2  —  xy  —  2y2-5x-2y  —  56  =  0. 

7.  2xz  —  xy  +  yz  —  7y  +  10  =  0. 
8. 

9. 

10.  4  cc2  —  2  xy  +  y2  —  4#  +  y  +  5  =  0. 

11.  2  a2  —  3xy  +  y2  —  6a,-  +  5?/  +  4  =  0. 

12.  Prove  that  the  general  equation  is  of  hyperbolic  type, 
if  AC<  0,  i.e.  if  -4  and  G  are  of  opposite  signs. 

13.  The  same,  if  B  =£  0  and  AC  =  0. 

5.   The  General  Equation,  B2  —  ±AC=  0.    .Fmrt  Method.    If 
jB2  —  4  ^4C  has  the  value  0,  the  equation 


(1) 

is  said  to  be  of  parabolic  type.  The  method  used  in  the  case 
.B2  —  4^4(7^=0,  which  begins  with  shifting  the  origin  so  that 
the  linear  terms  in  x  and  y  drop  out,  is  inapplicable  here,  since 
equations  (4)  of  §  4,  for  the  determination  of  the  new  origin, 
have  in  general  no  solution  if  jB2—  ±AC=  0.* 

Let  us  begin,  not  with  a  change  of  origin,  but  with  a  rota- 
tion of  axes,  assuming  that  B=fcO.  Applying  to  (1)  the  trans- 
formation (2)  of  §  2,  we  obtain 


(2)  ax'2  +  bx'y'  +  cy12  +  dx'  +  ey'  +  F=0, 

where  a,  6,  c  are  as  given  by  formulas  (3)  of  §  3,  and 

„  d=      Dcosy  +  .£/siny, 

e  =  —  D  sin  y  +  E  cos  y. 

*  They  have  no  solution  if  the  lines 

2  Ax  +  By  +  D  =  0,         Bx  +  2  Gy  +  E  =  0 
are  parallel  ;  infinitely  many  solutions,  if  these  lines  are  identical. 


251 

Since  formulas  (3)  of  §  3  are  valid,  so  are  the  equations  which 
were  deduced  from  them ;  in  particular, 

(4)  a  +  c  =  A+C, 

(5)  62-4ac  =  jB2-4^C. 

Here,,  B2  —  4:  AC  =  0,  and  hence  62  -  4  ac  =  0.  It  follows, 
then,  that  we  can  make  b  =  0  by  choosing  y  so  that  b  =  0,  or 
by  choosing  y  so  that  either  a  =  0  or  c  =  0.  The  second  of 
these  two  methods  is  in  the  end  the  simpler.  We  will  follow 
ill  and,  in  particular,  choose  to  make  a  =  0. 

If  a  =  0,  we  have,  by  the  first  of  the  formulas  (3)  of  §  3, 

A  cos2  y  +  B  sin  y  cos  y  +  C  sin2  y  =  0. 

TJ2 

Divide  by  sin2y,  substitute  for  Cits  value  — -,*  and  clear  of 

fractions ;  the  result  is 

4  A*  cot2  y  +  4:AB  cot  y  +  B2  =  0, 
or  (2  A  cot  y  +  B)2  =  0. 

Hence 

(6)  COty  — 

We  choose  that  value  of  y  satisfying  (6)  which  lies  between 
0°  and  180°. 

If  the  axes  are  rotated  through  this  angle  y,  then  a  =  0, 
6  =  0  and,  from  (4),  c  =  A  +  C.  Thus  (2)  becomes 

(7)  (A  +  C)y'2  +  dx'  +  ey'  +  F=0, 

where  the  values  of  d  and  e  are  to  be  computed  from  (3). 

Equation  (7)  can  now  be  treated  by  the  method  of  §  1,  Ex- 
ample 3. 

Example.     The  equation 

(8)  3z2  +  12o;?/  +  12?/2  +  10a  +  10?/-3  =  0    , 

is  of  parabolic  type,  since  B2  —  4  AC  =  144  —  4  •  3  •  12  =  0. 
*  A  =£  0,  for  otherwise  B  =  0  ;  and  we  have  assumed  B  ^  0. 


252 


ANALYTIC  GEOMETRY 


Here,  (6)  becomes 

whence 

y  =  153°  26', 

Rotate  the  axes  through  this  angle  y  and  compute  the  values 
which  A  +  C,  d,  and  e  have  in  this  case.  There  results,  as 
the  equation  into  which  (8)  transforms, 

15  y'2  -  2V5V  -  6V5V  -3  =  0. 
This  equation  can  be  rewritten  as 

or  15  (y'  -  \ V5>  =  2  V5  (x1  +  f  V5), 

or,  finally,  as 

(9) 
where 

(10) 

has  been  introduced  as  new  origin  of  coordinates. 

It  follows  from  (9)  that  equation  (8)  represents  a  parabola 
with  its  vertex  at  the  new  origin  (10)  and  with  its  axis  inclined 

at  an  angle  of  153°  26'  to  the  axis  of  x. 
To  find  the  coordinates  of  the 
vertex  (10)  with  respect  to  the 
original  axes,  we  substitute  in  for- 
mulas (2)  of  §  2,  first,  the  values 
which  sin  y  and  cos  y  have  in  this 

case  : 
/»•• 

FIG.  6 


V5 


V5 
We  obtain,  as 


and  then  the  values  for  x',  y'  given  by  (10). 
the  desired  coordinates,  (x,  ?/)  =  (!,  —  1). 

We  return  now  to  the  general  case.     If  d  =£  0,  equation  (7) 
represents  a  parabola  ;  cf.  Ex.  16,  §  1.     If  d  =  0,  (7)  can  be 
written  in  the  form 
(11)  y* 


EQUATION   OF  THE   SECOND   DEGREE          253 

where  we  have  divided  through  by  A  -f-  C  *  and  introduced 
simpler  notations  for  the  resulting  constants.  Equation  (11) 
becomes  immediately 

(12)  (y'  +  A02  =  fc2  +  *» 

consequently,  it  represents  two  parallel  lines,  a  single  line,  or 
has  no  locus,  according  as  k2  4-  1  is  positive,  zero,  or  negative. 

To  obtain  the  condition  for  the  exceptional  case,  d  =  0,  in 
terms  of  the  coefficients  of  (1),  we  note  that,  since  B2  —  4  AC 
=  0,  AC>  0  and  A  and  C  are  of  the  same  sign.  We  assume 
that  A  and  C  are  positive  ;  if  they  were  negative,  equation  (1) 
could  be  multiplied  through  by  —  1.  Since 

(13)  B  =  ±  2^  AC, 
(6)  can  be  written  as 


whence  it  can  be  shown  that 

C 


siny  = 
Hence,  from  (3), 


-VA+C  ^/A 

If  d  =  0, 
(14) 


Squaring  and  replacing  T  2^/AC  by  —  B,  we  have 
(15)  AE2  +  CD2  -  BDE  =  0. 

Now  A  may  be  written  as 

A  =  F(±AC  -  £2)  -  (AE2  +  CD2  -  BDE). 

Since  we  are  treating  the  case  B2  —  4  AC  =  0,  it  follows  from 
(15)  that  A  =  0 ;  conversely,  if  A  =  0,  then  (15),  and  hence 
(14),  holds  and  d  =  0.  Thus  the  condition  for  the  exceptional 
case  is  A  =  0. 

*  A  +  C  =£  0,  since  otherwise  (7),  and  therefore  (1),  would  not  be  of 
the  second  degree. 


254  ANALYTIC  GEOMETRY 

We  collect  our  results  in  the  form  of  a  theorem. 
THEOREM  5.     An  equation  (1)  of  parabolic  type: 
.B2- 4^0=0, 

represents  a  parabola,  (/"A  =£  0.    If  A  =  0,  it  represents  two  parallel 
lines,  a  single  line,  or  has  no  locus. 

We  have  proved  the  theorem  on  the  assumption  that  B  ^  0. 
If  B  =  0,  then  either  A  =  0  or  (7=0,  and  the  content  of  the 
theorem  is  easily  verified. 

If  A  =  0,  the  given  equation  may  be  treated  directly,  with- 
out change  of  axes.  An  equation  of  this  type  is 

8x2  +  24:xy  +18y2  —  14x  -  21y  +  3  =  0. 
It  can  be  written  in  the  form 

2(2x  +  3yy-  7(2 x  +  By)  +  3  =  0 ; 
the  left-hand  side  can  then  be  factored  : 

[2(2 aj  +  3y)  -  1]  [(2a?  +  3y)-  3]=  0. 
The  equation,  therefore,  represents  the  two  parallel  lines 
4a?  +  6y  — 1=0,         2x  +  3y-3  =  0. 

Second  Method.  We  notice  that  equation  (8)  can  be  written 
in  the  form 

(16)  3(x  +  2^)2  +  10a;  +  10y  —  3  =  0. 

There  are  two  linear  expressions  in  (16),  namely,  that  in  the 
parenthesis  and  that  consisting  of  the  remaining  terms  in  the 
equation.  If  the  lines  represented  by  these  expressions,  set 
equal  to  zero,  were  perpendicular,  (16)  could  be  simplified  by 
introducing  these  lines  as  coordinate  axes.  As  the  equation 
stands,  these  lines  are  not  perpendicular.  We  can,  however, 
rewrite  it  in  a  form  in  which  they  will  be. 

Add  an  arbitrary  constant  Tc  to  the  expression  in  the  paren- 
thesis in  (16)  and,  in  equalization,  subtract  6  k(x  +  2  y)  +  3  fc2 
from  the  remaining  terms : 

(17)  3(»  +  2y  +  fc)2+(10  -  6Jc)x  +  (10  -  12%  -  3  -  3fc2  =  0. 


EQUATION   OF  THE   SECOND   DEGREE 


255 


Determine  k  so  that  the  two  lines  defined  by  the  linear  ex- 
pressions in  (17)  are  perpendicular,  that  is,  so  that 


10  -12  A; 
Thus  -10  +  6fc  =  20  -24  k    and     fc=l, 

For  k  =  1,  (17)  becomes 
(18)  3(x  +  2  y  +  1)'  +  2(2  x  -  y  -  3)  =  0, 

and  the  two  lines  in  question  are  perpendicular.  The  equations 
of  a  transformation  introducing  these  lines  as  axes  were  found 
in  §  4  of  Ch.  XI,  and  are  given  by 
formulas  (5)  of  that  paragraph. 

Keferred    to    the    new   axes,    (18) 
becomes 


or 


FIG.  7 


Hence  we  have  shown  again  that 
(8)  represents  a  parabola  with  its  vertex  at  the  point  (1,  —  1) 
and  with  its  axis  inclined  at  an  angle  of  153°  26'  to  the  axis 
of  x* 

The  first  of  the  two  methods  described  is  more  direct  and 
more  in  keeping  with  previous  methods.  Its  application  to 
a  particular  equation,  however,  is  handicapped  by  the  early 

*To  treat  the  general  equation  (1)  by  this  method,  assume  that  A  and 
C  are  positive.  Then,  since  B  =  ±  2  VJ.C,  (1)  can  be  written  as 

(  VAx  ±  VCy)*  +  Dx  +  Ey  +  F  =  0. 

From  this  point  the  discussion  proceeds  as  in  the  example  in  the  text. 
It  can  be  shown  that  the  exceptional  case  arises  when  and  only  when 

V2x  ±  VCy  =  0,         Dx  +  Ey  +  F  =  0 
are  parallel  ;  i.e.  when  and  only  when 

EVA  T  D  VC  =  0. 

But  this  is  precisely  the  equation  (14)  obtained  by  the  first  method. 
From  it  follows  that  A  =  0  is  the  condition  for  the  exceptional  case. 
Thus  we  have,  in  sketch,  the  proof  of  Theorem  5  by  the  second  method. 


256 


ANALYTIC   GEOMETRY 


introduction  of  radicals.  The  second  method  avoids  this  dis- 
advantage, and  is  the  more  elegant,  though  perhaps  theoreti- 
cally the  more  difficult,  of  the  two. 

EXERCISES 

Identify  and  plot  roughly  the  curve  denned  by  each  of  the 
following  equations.  If  a  change  of  axes  is  necessary,  show 
the  new  axes  on  the  graph. 


2.  9z2- 

3.  25z2  +  120ary  +  144  y*  +  86  x-233y  +  270  =  0. 

4.  5x2-20xy  +  2Qy2  +  2x  +  y  +  3=  0. 

5.  25x*  +  30xy  +  9y2+  Wx  +  6y+l  =  Q. 

6.  a2  —  2  xy  +  f-  +  3  x  —  y  —  4  =  0. 

7.  x"<-4:xy  +  4:y*  +  3x—  6#-10  =  0. 

8.  27  x2  -  36xy  +  12y*  —  40z  +  18  y  +  32  =  0. 

9.  2xz  +  12xy  +18i/24-a;  +  137/  +  9  =  0. 

10.  4z2  -f  12z?/  +  92  +  2x  +  3i/  +  2  =  0.     ^Ins.     No  locus. 


6.   Summary.     Invariants.     The  content  of  Theorems  3,  4,  5 
we  summarize  in  the  following  table. 


JB2_4^<7<0 

J32-  4AC  =  0 

52-4^00 

A  =£0 

Ellipse,  if  AA.  <  0 
No  locus,  if  J.A  >  0 

Parabola 

Hyperbola 

A  =0 

Point 
(Null  Ellipse) 

Two  parallel  lines, 
a  single  line,  or 
no  locus 

Two  intersecting 
lines 

The  ellipse,  parabola,  and  hyperbola  are  plane  sections  of  a 
right  circular  cone  (Ch.  VIII,  §  10).  Now,  the  section  of  the 
cone  by  a  plane  through  the  vertex  is  a  point,  a  single  line,  or 
two  intersecting  lines.  If  the  vertex  of  the  cone  is  carried 
off  in  the  direction  of  the  axis  indefinitely,  the  cone  approaches 


EQUATION  OF  THE  SECOND  DEGREE  257 

a  cylinder  as  its  limit,  and  the  plane  approaches  a  position 
parallel  to  the  rulings  of  the  cylinder.  But  the  section  of  a 
cylinder  by  a  plane  parallel  to  the  rulings  is  two  parallel  lines, 
a  single  line,  or  nothing.  These  sections  of  the  cone  or  cylin- 
der are  called  degenerate ;  those  first  mentioned,  non-degenerate. 
From  the  above  table  we  can  now  draw  a  general  conclusion. 

THEOREM  6.  An  equation  of  the  second  degree,  if  it  has  a 
locus,  represents  a  conic  section,  ivhich  is  non-degenerate  if  A  =£  0, 
and  degenerate  if  A  =  0. 

Invariants.  We  have  seen  that  the  value  of  the  quantity 
A  +  C  is  unchanged  by  a  rotation  of  axes  [§  3,  (7)  and  §  5,  (4)]. 
This  is  true  also  of  the  value  of  the  quantity  ZJ2  —  4.4(7  [§  3, 
(16)  and  §  5,  (5)].  We  say  that  A  +  C  and  W-±AC  are 
invariant  under  a  rotation  of  axes.  They  are  ajso  invariant 
under  a  change  of  origin,  since  we  saw,  in  §  4,  that  the  quadratic 
terms  in  the  general  equation  are  not  affected  by  a  change  of 
origin. 

Consequently,  A  +  C  and  B2  —  4  AC  are  invariant  under  any 
change  of  axes.  For,  any  change  of  axes  consists  of  a  change 
of  origin,  combined  with  a  rotation  of  axes. 

It  can  be  shown  that  the  discriminant  A  is  also  invariant 
under  any  change  of  axes. 

The  importance  which  these  quantities,  A,  W  —  4.4(7,  and 
A  +  C,  have  assumed  in  the  course  of  the  treatment  is  closely 
related  to  the  fact  that  they  are  invariants  with  respect  to  any 
change  of,  axes.  For,  it  is  clear  that  a  quantity  whose  value 
varies  with  the  choice  of  axes  can  have  no  particular  signifi- 
cance in  a  theory  which  deals  primarily  with  properties  of 
the  curve  which  are  independent  of  the  choice  of  axes,  whereas 
it  is  to  be  expected  that  an  invariant  quantity  would  play  an 
important  role. 

EXERCISES  ON  CHAPTER  XII 

In  each  of  the  following  exercises,  determine  the  nature  of 
the  curve  represented  by  the  given  equation,  and  then  find  its 


258  ANALYTIC   GEOMETRY 

position.     Draw  a  figure,  showing  the  curve,  the  original  axes, 
and  any  new  axes  used. 

1.  Ilx2  +  6xy  +  3y*-l2x-l2y -12  =0. 

2.  7 x2  -  8 xy  +  ?/2  +  14 a  -  8.y  +  16  =  0. 

3.  8z2  +  8z?/  +  22/2-6a;-32/-5  =  0. 

4.  4x2  +  8>y  +  4t/2  +  13z  +  3y  +  4  =  0. 

5.  9a2-8o*/  +  24?/2-32a;-16?/  +  138  =  0. 

6.  a;2  +  ccy  -  2y2  -  11  a:  -  y  +  28  =  0. 
7. 

8. 

9. 
10. 
11. 
12. 
13. 
14. 
15. 

16.  7a?  —  18xy  —  17if-  -28x  +  36y  +  8  =  0. 

17.  9x2- 12a;y  +  4y2  +  4a;-59i/  +  38  =  0. 

18.  7x*—5xy  +  y*  —  42«  +  15y +  63  =  0. 

19.  14a;2  +  2±xy  +  2ly*  +  52x  +  66  y  +  14  =  0. 
20. 

21. 

22.  25x2-7x   +   z— 107 x+W    + 13  =  0. 

23.  49x2 

24.  4a;2 

25.  a;2  — 

26.  2  a2  —  xy  +  y2  —  7  ?/  +  6  =  0. 

27.  Show  that  an  equation  of  the  second  degree  represents 
an  equilateral  hyperbola  or  two  perpendicular  lines,  if  and 
only  ifA  +  C  —  0. 


EQUATION  OF  THE  SECOND  DEGREE  259 

28.  If  the  equation  (1),  §  3,  represents  a  hyperbola,  show 
that  the  asymptotes  are  defined  by  the  equation 

Ax*  +  Bxy  +  Cy2  =  0. 

29.  If  the  general  equation  of  the  second  degree  represents 
a  hyperbola,  prove  that  the  asymptotes  have  the  directions  of 
the  lines  defined  by  the  equation  of  Ex.  28. 

30.  Prove  that  every  equation  of  the  form 

AB  (x*  -  y2)  -  (A*  -  B*)xy  =  C, 

where  C=£0  and  not  both  A  and  B  are  0,  represents  a  rectan- 
gular hyperbola  with  the  lines 

Ax  +  By  =  0,  Bx  —  Ay  =  Q 

as  asymptotes. 

31.  Show  that  the  equation  of  every  rectangular  hyperbola 
can  be  written  in  the  form 


AB(xi  -  y*)-(A*  -  B*)xy  +  Dx  +  Ey  +  F=  0. 

32.  Find  the  equation  of  each  of  the  rectangular  hyperbolas 
(a)  12x*-7xy-12y'*-17x  +  31y-13  =  0, 

(6)  6x2  +  5xy-6y*-39x  +  26^-13  =  0, 

referred  to  the  asymptotes  as  axes. 

33.  Show  that  for  just  one  value  of  A  the  equation 

Xa;2  +  4  xy  +  y*  —  4x  —  2  y  —  3  =  0 
represents  two  straight  lines.     Find  the  equations  of  the  lines. 

34.  If  the  general  equation  of  the  second  degree  represents 
an  ellipse  or  hyperbola,  what  is  the  condition  that  the  center 
be  at  the  origin  ? 

35.'  If  the  general  equation  represents  a  parabola,  show  that 
the  vertex  is  at  the  origin  if  and  only  if 

AD"-  +  CE*  +  BDE  =0  and  ^=0. 

Suggestion.     Write  the  equation  in  the  form  (7),  §  5. 


260  ANALYTIC   GEOMETRY 

36.  Prove  that,  if  the  equation  (1),  §  3,  represents  an  ellipse 
or  hyperbola,  the  axes  are  denned  by  the  equation 

B(x*  -  jf)  -  2  (A  -C)xy  =  0. 

37.  Show  that,  in  case  .B2  —  4  AC  =  0  and  A  =  0,  the  gen- 
eral equation  represents  two  parallel  lines,  a  single  line,  or 
has  no  locus,  according  as  the  expression 


is  positive,  zero,  or  negative. 

Suggestion.  Consider  equation  (7),  §  5,  where  d  =  0,  and 
use  (15),  §  5,  in  simplifying  the  result. 

38.  Prove  that  the  expression  in  Ex.  37  can  be  replaced  by 
D2  -  4  AF,  if  A  3=.  0  ;  and  by  E*  -  4  OF,  if  C=?=  0. 

Definition.  Two  conies  are  said  to  be  similar  and  similarly 
placed,  if  their  eccentricities  are  equal  and  their  corresponding 
axes  are  parallel. 

39.  Prove  that  the  conies, 

11  xz  -f  6  xy  +  3  y2  —  12  x  —  12y  -  12  =  0, 
11  x2  +  6xy  +  3y*  -  34  a;  -  ISy  +  29  =  0, 
are  similar  and  similarly  placed. 

40.  Show  that,  if  the  coefficients  of  the  quadratic  terms  in 
two  equations  which  represent  nonrdegenerate  conies  are  re- 
spectively equal  or  proportional,  the  conies  are  of  the  same 
type.     Prove  further  that,  if  they  are  ellipses  or  parabolas, 
they  are  similar  and  similarly  placed,  and  that,  if  they  are 
hyperbolas,  they  are  similar  and  similarly  placed  or  each  is 
similar  and  similarly  placed  to  the  conjugate  of  the  other. 


CHAPTER   XIII 


A   SECOND   CHAPTER    ON  LOCI.     AUXILIARY  VARIABLES. 
INEQUALITIES 

1.  Extension  of  the  Method  for  the  Determination  of  Loci. 

If  we  look  back  over  the  locus  problems  which  we  have  thus 
far  solved,  we  find  that  there  is  invariably  but  a  single  essen- 
tial condition  governing  the  motion  of  the  point  tracing  the 
locus.  For  example,  the  sum  or  the  difference  of  two  dis- 
tances or  of  their  squares  is  required  to  be  .constant ;  or  the 
slope  of  one  line  is  given  as  proportional  to  that  of  another. 
This  single  condition  it  is  comparatively  simple  to  express 
analytically  and  thus  to  determine  the  locus. 

In  a  great  many  problems,  however,  the  motion  is  governed 
by  not  just  one,  but  by  two  or  more  essential  conditions,  in- 
terdependent on  one  another. 
An  example  of  such  a  problem 
is  the  following :  A  triangle 
has  a  fixed  base  AB  and  its 
vertex  V  moves  on  an  indefinite 
straight  line  L,  parallel  to  the 
base.  Find  the  locus  of  the 
point  of  intersection  P  of  the  ~v .  /_a  Q\ 
altitudes. 

Here    the    motion    of    P   is 

governed  by  tivo  essential  conditions ;  first,  by  the  motion  of 
V,  and  secondly,  by  the  fact  that  P  is  the  point  of  intersection 
of  the  altitudes.  The  two  conditions  are  interrelated,  since 
the  position  of  V  determines  the  position  of  the  altitudes  and 
hence  of  their  point  of  intersection.  Consequently,  they  are 

261 


V:(y,h) 


O      C  B:(a,0) 
FIG.  1 


262  ANALYTIC   GEOMETRY 

in  substance  equivalent  to  a  single  condition,  which,  when 
expressed  analytically,  would  give  the  equation  of  the 
locus. 

Our  problem,  then,  is  to  reduce  the  two  conditions  to  a  single 
condition.  This  it  is,  in  general,  very  difficult  to  do  geometri- 
cally. Analytically,  however,  the  task  is  simpler.  For,  condi- 
tions expressed  in  analytical  form,  in  terms  of  equations,  are 
usually  more  easily  combined  than  when  they  are  in  geometri- 
cal form. 

Accordingly,  we  proceed  to  express  analytically  the  two 
conditions  governing  the  motion  of  P.  Take  the  mid- 
point of  the  base  of  the  triangle  as  the  origin  and  the 
axis  of  x  along  the  base.  Let  the  length  of  the  base  be 
2  a  and  the  distance  of  L  above  the  base,  h.  '  The  coordinates 
of  A  and  B  are  (—  a,  0)  and  (a,  0).  Denote  those  of  P  by 

(X,  Y}. 

The  first  of  the  two  conditions  is  that  V  move  along  L. 
But  then  the  distance,  KV,  of  V  from  the  axis  of  y  varies. 
Accordingly,  we  can  express  the  motion  of  V  along  L  by  tak- 
ing the  abscissa  of  F"as  a  variable.  Denote  this  variable  by  y. 
The  coordinates  of  Fare,  then,  (y,  h). 

We  now  have  coordinates  for  the  three  vertices  of  the  tri- 
angle. Hence  we  can  find  the  coordinates  (X,  Y)  of  the 
point  of  intersection  of  the  altitudes.  Thus  we  shall  have 
expressed  the  condition  that  P  be  this  point. 

The  coordinates  (X,  Y)  of  P  will  be  obtained  in  terms  of 
the  constants,  a  and  h,  and  the  variable  y.  If  we  eliminate  y 
from  the  two  equations  which  give  the  values  of  X  and  Y,  the 
resulting  equation  will  contain  only  a  and  h,  and  X  and  Y,  and 
will  be  the  equation  of  the  locus  of  P. 

The  variable  y  is  known  as  an  auxiliary  variable,  or  param- 
eter. It  helps  in  expressing  analytically  the  conditions 
governing  the  generation  of  the  locus.  The  method  involving 
its  use,  which  we  have  just  described,  is  general  in  scope,  and 
may  be  applied  with  advantage  to  any  locus  problem  contain- 
ing multiple  conditions. 


A    SECOND    CHAPTER   ON  LOCI  263 

2.  One  Auxiliary  Variable.  In  illustrating  the  method  by 
examples,  let  us  first  complete  the  problem  of  the  previous 
paragraph. 

Consider  P  as  the  point  of  intersection  of  VC  and  BD. 
The  equation  of  VC  is 

(1)  X=y. 

The  slope  of  A  V  is 

h 

y  +  «' 
hence  the  equation  of  the  perpendicular,  BD,  to  A  V  is 

(2) 

Solving  equations  (1)  and  (2)  simultaneously,  we  obtain  the 
coordinates  of  P, 

(3)  X=y,  Y=°> 


in  terms  of  the  constants  a  and  h  and  the  auxiliary  variable  y. 
By  eliminating  y  from  equations  (3),  we  obtain 

(4)  X*  =  -hY+a? 

as  the  equation  of  the  locus. 

The  locus  of  P  is,  then,  a  parabola,  with  its  axis  along  the 
perpendicular  bisector  of  the  base  of  the  triangle  ;  it  goes 
through  the  extremities  of  the  base  and  opens  away  from  the 
line  L.  Every  point  of  it  is  included  in  the  locus.* 

*  In  the  locus  problems  considered  hitherto,  particularly  in  Ch.  V, 
care  was  taken  to  emphasize  that  two  things  are  necessary  :  (a)  to 
determine  the  curve,  or  curves,  on  which  points  of  the  locus  lie;  (6)  to 
show,  conversely,  that  eVery  point  lying  on  the  curve,  or  curves,  obtained 
is  a  point  of  the  locus.  In  the  problems  of  the  present  chapter,  —  for 
example,  in  the  one  above,  —  part  (6)  of  the  proof  is  usually  omitted.  It 
consists,  as  a  rule,  in  retracing  the  steps  of  part  (a)  and  so  presents,  in 
general,  no  difficulty.  And  it  is  more  important,  now,  that  the  student 
gain  facility  in  deducing  the  equation  of  the  curve,  or  the  equations  of  the 
curves,  which  turn  out,  in  the  great  majority  of  cases,  to  be  precisely  the 
locus. 


264 


ANALYTIC   GEOMETRY 


Remark.  It  was  not  necessary  to  find  the  actual  coordinates 
(3)  of  P.  The  fact  that  P  is  the  point  of  intersection  of  the 
altitudes  might  have  been  expressed  by  writing  down  the 
conditions  that  (X,  F)  satisfy  equations  (1)  and  (2),  namely, 


If  we  eliminate  y  from  these  equations,  we  obtain  equation 
(4)  of  the  locus. 

Example  2.     A  straight  line  L  passes  through  a  fixed  point 
P0 ;  find  the  locus  of  the  mid-point  P  of  the  portion  of  L  inter- 
cepted by  two  given  perpendicu- 
lar lines,  neither  of  which  goes 
through  P0. 

Take  the  two  given  lines  as 
axes,  and  let  the  coordinates  of 
P0,  referred  to  them;  be  (x0,  y0)- 
The  conditions  governing  the 
motion  of  P  are,  first,  the  rota- 
tion of  L  about  P0,  and  secondly, 
the  fact  that  P  is  the  mid-point 
of  the  segment  AB. 
.  We  express  the  rotation  of  L 
by  taking  its  slope,  X,  as  aux- 


FIG.  2 


iliary  variable.     The  equation  of  L  is,  then, 

The  coordinates  of  the  points  of  intersection  of  L  with  the 


axes  are : 


Hence  the  coordinates  of  P,  the  mid-point  of  AB,  are 


To  eliminate  A  from  these  two  equations,  we  might  solve 
the  first  for  X  and  substitute  its  value  in  the  second.     But  we 


A   SECOND   CHAPTER  ON  LOCI  265 

notice  an  easier  method  ;  rewriting  the  equations  in  the  form 

O    V-  rf      _  2/0 

*  -A-  —  x0  ---  -  , 
A 

2  T—y0  =  -Xx0, 

and  multiplying  together  the   left-hand  sides   and  then  the 
right-hand  sides,  we  obtain  the  equation 

(5)  (2X-x0)(2T-y0)=x0y0) 

devoid  of  X. 

The  equation  of  the  locus,  in  this  form,  or  better,  in  the 
form: 


suggests  that  we  change  to  parallel  axes,  with  the  new  origin 


-, 

The  locus,  referred  to  the  new  axes,  has  the  equation, 


and  is,  therefore,  a  rectangular  hyperbola.     It  follows  from  (5) 
that  the  hyperbola  goes  through  0  and  P0. 

To  describe  the  locus  independently  of  the  coordinate  sys- 
tem :  Let  0  be  the  point  of  intersection  of  the  given  lines  ; 
the  locus  is  a  rectangular  hyperbola  through  0  and  P0,  with 
its  center  at  the  mid-point  of  OP0  an(i  with  its  asymptotes 
parallel  to  the  given  lines. 

EXERCISES 

1.  Given  a  line  L  parallel  to  the  axis  of  x.  Through  the 
origin  draw  a  variable  line  meeting  L  in  Q,  and  on  this  vari- 
able line  mark  the  point  P  whose  ordinate  equals  the  abscissa 
of  Q.  What  is  the  locus  of  P  ? 

Ans.  The  parabola  y"1  =  hx,  where  h  is  the  algebraic  distance 
from  the  axis  of  x  to  L. 


266  ANALYTIC   GEOMETRY 

2.  A  line-segment  AB  of  fixed  length  moves  so  that  its 
extremities  lie  always  on  two  perpendicular  lines.  Find  the 
locus  of  the  point  dividing  AB  in  the  ratio  2 : 1,  using  as 
auxiliary  variable  the  angle  which  the  moving  line  makes  with 
one  of  the  two  perpendicular  lines. 

Ans.  An  ellipse,  center  in  the  point  of  intersection  of  the 
given  lines,  axes  along  them,  with  length  and  breadth  in  the 
ratio  2  : 1. 

3.  Determine  the  locus  described  in  Ex.  2,  when  the  given 
ratio  is  mj :  m2. 

4.  The  line  L  is  the  perpendicular  bisector  of  the  fixed 
horizontal  line-segment  AB.     The  points  R  and  S  are  taken 
on  L,  with  R  always  below  S,  so  that  the  distance  US  is  one 
half  the  distance  AB.     Find  the  locus  of  the  point  of  inter- 
section of  AR  and  BS,  taking  the  axes  of  x  and  y  along  AB 
and  L.     Ans.   The  hyperbola,  2  xy  -f  a?  =  a2,  through  A  and  B. 

5.  A  variable  line  is  drawn  through  a  fixed  point  Pl  meet- 
ing a  fixed  line  L  in  P2.     Points  P  are  taken  on  this  line  so 
that  the  product  of  the  distances  P^  and  PiP-2  is  constant. 
Find  the  locus  of  these  points. 

Ans.  Two  circles,  tangent  at  PI  to  the  line  through  P^  par- 
allel to  L. 

6.  Find  the  locus  of  points  from  which  the  tangents  drawn 
to  a  parabola  are  perpendicular. 

Suggestion.  Use  the  equation  of  the  tangent  with  given 
slope,  Ch.  IX,  §  6,  eq.  (10). 

7.  The  same  for  the  ellipse. 

8.  The  same  for  the  hyperbola. 

9.  Determine  the  locus  of  the  mid-points  of  all  the  chorda 
drawn  from  the  vertex  of  a  parabola. 

3.  Coordinates  of  a  Point  Tracing  a  Curve,  as  Auxiliary  Vari- 
ables. In  the  problems  of  the  previous  paragraph  one  of  the 
conditions  governing  the  motion  of  the  point  tracing  the  locus 
was  the  auxiliary  motion  of  a  line  or  of  a  second  point.  In 


A   SECOND    CHAPTER  ON   LOCI  267 

each  case  this  auxiliary  motion  could  be  expressed  analytically 
by  the  introduction  of  one  auxiliary  variable. 

Suppose,  now,  that  the  auxiliary  motion  consists  of  the  trac- 
ing of  a  given  curve,  not  a  straight  line,  by  a  point  R.  Let 
the  curve  be,  for  example,  the  circle 

(1)  »2  +  y~  =  a2. 

The  motion  of  R  on  the  circle  might  be  represented  analyti- 
cally by  the  introduction  of  a  single  auxiliary  variable,  e.g. 
one  of  the  coordinates  of  R ;  but  it  is  in  general  simpler,  ana- 
lytically, to  represent  the  motion  by  two  auxiliary  variables, 
namely,  by  both  the  coordinates  (xf,  y'}  of  R.  These  will  be 
connected  by  the  equation, 

(2)  a^  +  y2  =  a2, 

which  states  that  R  is  on  the  circle. 

The  reason  for  this  choice  of  auxiliary  variables  lies  partly 
in  the  fact  that  we  thereby  avoid  radicals ;  *  partly  in  the 
principle  of  algebraic  symmetry.  By  this  term  we  mean  to 
signalize  the  fact  that  equation  (1)  bears  equally  on  x  and  y, 
and  so  it  is  well  to  carry  the  solution  through  in  such  a  man- 
ner that  it,  too,  will  bear  equally  on  the  two  coordinates  of 
each  of  the  principal  points  involved. | 

Example  1.  Let  AA!  be  a  fixed  diameter  of  a  given  circle 
and  let  RR'  be  a  variable  chord  perpendicular  to  AA'.  What 
is  the  locus  of  the  point  of  intersection,  P,  of  AR  and  A'R'  ? 

Choose  the  center  of  the  circle  as  the  origin  and  the  axis  of 
x  along  AA'.  Then  (1)  is  the  equation  of  the  circle. 

*  If  we  had  taken  x'  as  a  single  auxiliary  variable,  the  coordinates  of 
R  would  be  (x',  ±  Va2  — x'2). 

t  It  is  possible  to  represent  the  motion  of  -R  by  a  single  auxiliary  vari- 
able and  at  the  same  time  to  avoid  radicals  and  preserve  symmetry,  by 
choosing  as  the  auxiliary  variable  the  angle  6  which  the  radius  drawn  to 
R  makes  with  a  fixed  direction,  e.g.  the  axis  of  x  ;  the  coordinates  of  jR 
are  then  :  x  —  a  cos  0,  y'  =  a  sin  6.  We  prefer,  however,  to  use  as  aux- 
iliary variables  the  coordinates  of  R  connected  by  equation  (2). 


268 


ANALYTIC   GEOMETRY 


Take,  as  the  auxiliary  motion,  the  tracing  of  the  circle  by 
the  point  R  and,  as  auxiliary  variables,  the  coordinates  (#',  y'} 

of  R,     These  are  connected 
v  by  the  equation  (2).     The 

(X  Y)    coordinates  of  R'  are,  evi- 
dently, («',  -  y'\ 

The    equations    of    AR 
and  A'R'  are 


x  —  a 


FIG.  3 


its  coordinates  (X, 
(3) 

(4) 


Since  P  is  the  point  of  in- 
tersection of  AR  and  A'R', 
satisfy  both  these  equations : 


x'  —  a 


We  have,  then,  three  equations,  (2),  (3),  and  (4),  involving, 
besides  the  constant  a,  the  coordinates  (X,  Y)  of  the  moving 
point  and  the  auxiliary  variables  x',  y'.  To  obtain  an  equa- 
tion in  X,  y  alone,  we  must  eliminate  x',  y'.^  We  shall  do  this 
by  solving  two  of  these  equations,  preferably  (3)  and  (4), 
simultaneously  for  x'  and  y'}  and  substituting  the  values 
obtained  for  them  in  the  third  equation,  (2). 

To  this  end  we  rewrite  equations  (3)  and  (4)  as  follows : 
(3a)  Yxf—(X+a~)yf=-aY, 

(4a)  Yx'+(X-a}y'  =      aY. 

Hence  x'  =  —  •,  y'  =  a — 

Substituting  these  values  in  (2)  and  reducing,  we  obtain 

X2  -  F2  =  a2 
as  the  equation  of  the  locus. 


A   SECOND   CHAPTER  ON   LOCI 


269 


The  locus  is  thus  seen  to  be  a  rectangular  hyperbola  with  the 
given  diameter  of  the  circle  as  major  axis.  It  is  evident  from 
the  figure,  however,  that  if  R  is  restricted  to  the  upper  half  of 
the  circle,  and  R'  to  the  lower  half,  only  the  upper  half  of  one 
branch  and  the  lower  half  of  the  other  belong  to  the  locus.  It 
is  only  when  R  and  R'  are  each  permitted  to  trace  both  halves 
of  the  circle  that  the  locus  consists  of  the  entire  hyperbola. 

Remark.  The  points  A  and  A'  do  not  belong  to  the  locus. 
For,  the  only  possible  way  in  which  P  can  take  on  the  position 
of  the  point  A',  for  example,  is  for  R  and  R'  to  coincide  in  A'  ; 
but  then  there  is  no  chord  RR1  and  also  no  line  A'R',  so  that 
no  point  P  on  the  locus  is  determined. 

Let  us  return  now  to  equations  (3a)  and  (4a).  In  solving 
them  for  x',  we  actually  obtain 

TXx'  =  a?  Y 


But  Y"=£  0,  since  P  cannot  lie  on  the  axis  of  x,  in  either  A  or 
A'  ;  hence  we  were  justified  in  dividing  by  Y,  and  the  result, 
x'  =  a?/X,  is  correct. 

In  subsequent  problems  we  shall  lay  no  stress  on  exceptional 
points  such  as  A  and  A'.  Their  importance  for  the  student  at 
this  stage  is  relatively  small. 

Example  2.  A  point  R  traces  a  pa- 
rabola. Find  the  locus  of  the  point  of 
intersection,  P,  of  the  line  through  the 
focus  and  R  with  the  line  through  the 
vertex  perpendicular  to  the  tangent 
at  R. 

The  parabola,  referred  to  the  coor- 
dinate axes  shown  in  the  figure,  has 
the  equation 

2/2  =  2  mx. 

The  motion  of  R  can  be  expressed  by  taking,  as  auxiliary 
variables,  its  coordinates  (as',  y'\  connected  by  the  relation 
(5)  y"y- 


FIG.  4 


270  ANALYTIC   GEOMETRY 

which   states   that  R,  in   moving,   stays   always   on   the   pa- 
rabola. 

The  slope  of  the  tangent  at  R  is  m/y',  Ch.  IX,  §  2,  eq.  (5); 
consequently,  the  line  through  0  perpendicular  to  the  tangent  is 

y' 

m 
As  the  equation  of  FR  we  have 

y    ( 
y=-^(*-o 


The  equations  expressing  the  fact  that  P :  (X,   F)  is  the 
point  of  intersection  of  these  two  lines  are,  therefore, 

(6)  Y=-^X, 


From  equations  (5),  (6),  and  (7)  we  have  to  eliminate  x'  and 
y'.  Solving  (6)  and  (7)  for  x'  and  y',  we  have : 

,  T  ,      m(m  —  X) 

y'=—m — ,  x' = —± 1, 

X'  2  X 

Substituting  these  values  for  y'  and  x'  in  (5)  and  reducing  the 
result,  we  obtain 

X2  +  F2  -  mX  =  0 

as  the  equation  of  the  locus. 

The  locus  is  therefore  a  circle,  passing  through  the  vertex  of 
the  parabola  and  having  its  center  at  the  focus.  The  vertex, 
0,  is  not  a  point  of  the  locus.* 

Elimination  of  x',  y'.  In  each  of  the  above  examples  we 
eliminated  the  auxiliary  variables  x',  y'  by  solving  the  last 
two  of  a  set  of  three  equations  for  x',  y'  and  substituting  the 
values  thus  obtained  for  x',  y'  in  the  first  equation,  —  the 

*  It  is  an  exceptional  point,  similar  in  type  to  the  exceptional  points, 
A  and  A',  of  Example  1.  For,  when  R  is  at  0,  FR  and  OP  coincide  and 
consequently  determine  no  point  on  the  locus. 


A   SECOND   CHAPTER  ON  LOCI  271 

equation  stating  that  the  point  (xr,  y')  lies  on  the  given  curve. 
This  method  is  valuable  because  of  its  general  applicability. 
The  student  should,  however,  be  on  the  alert  for  short  cuts  in 
the  elimination.  For  example,  he  might  have  noticed  by  close 
inspection  that,  in  Example  1,  x',  y'  can  be  eliminated  easily 
from  equations  (2),  (3),  (4)  by  multiplying  equations  (3)  and 
(4)  together  : 


and  by  noting,  from  equation  (2),  that  the  quantity 

-y'2 

x'2  -  a2 
has  unity  as  its  value. 

EXERCISES 

1.  Let  A  A'  be  the  major  axis  of  an  ellipse  and  RR'  be  a 
variable  chord  perpendicular  to  AA'.     Find  the  locus  of  the 
point  of  intersection  of  AR  and  A'R'. 

2.  Given  a  fixed  diameter  of  a  circle  and  a  variable  chord 
parallel  to  it.     Find  the  locus  of  the  point  of  intersection  of 
the  line  through  the  mid-point  of  the  chord  and  one  extremity 
of  the  diameter  with  the  radius  drawn  to  the  corresponding 
extremity  of  the  chord.     What  is  the  locus  if  the  radius  is 
drawn  to  either  extremity  of  the  chord  ? 

Ans.     Part  of  a  parabola  ;  the  parabola. 

3.  Find  the  locus  of  the  point  of  intersection  of  the  line 
drawn  through  a  given  focus  of  an  ellipse  perpendicular  to  a 
variable  tangent  with  the  line  joining  the  center  to  the  point  of 
tangency. 

Ans.     The  directrix  corresponding  to  the  given  focus. 

4.  Let  R  be  a  point  tracing  an  ellipse.     Find  the  locus  of 
the  point  of  intersection  of  the  line  drawn  through  the  center 
perpendicular  to  the  tangent  at  R  with  the  line  drawn  through 
R  parallel  to  the  conjugate  axis. 

Ans.     An  ellipse,  similar  to  and  having  the  same  axes  as 
the  given  ellipse,  but  with  foci  on  the  opposite  axis. 


272  ANALYTIC   GEOMETRY 

5.  The  preceding  problem  for  a  hyperbola. 

6.  Find  the  locus  of  the  point  of  intersection  of  the  line 
drawn  through  a  variable  point  R  of  a  parabola  parallel  to  the 
axis  and  the  line  through  the  vertex  perpendicular  to  the  tan- 
gent at  R. 

7.  A  variable  tangent  to  an  ellipse  meets  the  transverse 
axis  in  the  point  T.     Determine   the  locus  of   the  point   of 
intersection  of  the  line  drawn  through  T  parallel  to  the  con- 
jugate axis  and  the  line  joining  the  point  of  contact  of  the 
tangent  to  a  vertex. 

8.  The  preceding  problem  for  a  hyperbola. 

9.  Let  RR'  be  an  arbitrary  chord  of  an  ellipse  parallel  to 
the  conjugate  axis  ;  let  the  normal  at  R  meet  the  line  joining 
the  center  to  R'  in  the  point  S.     Find  the  locus  of  the  mid- 
point of  RS. 

10.    The  preceding  problem  for  a  hyperbola. 

4.  Other  Problems  Involving  Two  or  More  Auxiliary  Vari- 
ables. There  are  problems  in  which  it  is  convenient  to  use 
two  auxiliary  variables  other  than  those  of  the  type  which 
we  considered  in  the  preceding  paragraph. 

Example.  The  points  A  and  B  are  fixed  and  the  line  L  is 
perpendicular  to  AB  at  its  mid-point,  0;  R  and  S  are  two 

points  on  L,  both  on  the  same  side 
of   AB   and    moving    so    that  the 
product  of  their  distances  from  0 
P:(X,Y)        is  constant,  and  equal  to  V1.     Find 


(0,r)- 


A'-(-a,0) 


of  intersec- 


tion,  P,  of  AR  and  BS. 


~0  B'-  (a,0)       Take  the  axes  as  shown  in  the 


5  figure    and    let    AB  =  2  a.       The 

motions  of  R  and  S  can  be  repre- 

sented by  taking  their  ordinates,  which  we  denote  by  r  and  s, 
as  auxiliary  variables.     The  condition  that  R  and  S  are  on  the 


A   SECOND   CHAPTER   ON   LOCI  273 

same  side  of  AB  and  relatively  so  situated  that  OR  •  OS  =  b2 
is  then  given  by  the  equation, 
(1)  rs  =  b2. 

The  equations  of  AR  and  BS  are 


—  a     r  as 


Since  P  :  (X,  F)  is  the  point  of  intersection  of  these  lines,  we 
have 


To  eliminate  the  auxiliary  variables  r  and  s  from  equations 
(1),  (2),  (3)  is  now  our  problem.  We  notice  that  in  the  product 
of  equations  (2)  and  (3)  : 

i_.x2=Z! 

o2       rs  ' 

r  and  s  enter  only  in  the  form  rs,  and  that  the  value  of  rs  is 
given  by  (1)  as  &-.  We  have,  therefore,  as  the  equation  of  the 
locus 


The  locus  of  P  is,  therefore,  an  ellipse,  with  its  axes  along 
AB  and  L,  and  passing  through  the  points  A  and  B.  These 
points  are  not,  however,  points  of  the  locus. 

EXERCISES 

1.  What  is  the  locus  of  P  in  the  problem  in  the  text,  if  R 
and  S  are  always  on  opposite  sides  of  AB  ? 

2.  The  points  Px  and  P2  are  fixed,  and  the  lines  LI  and  L2 
are  perpendicular  to  PxP2  in  P:  and  P2  respectively  ;  Qi  and 
Q.2  are  two  points  on  LI  and  L2  respectively,  both  on  the  same 
side  of  PiP2  and  mpving  so  that  the  product  of  their  distances 


274  ANALYTIC  GEOMETRY 

from  P1  and  P2  respectively  is  constant.     Find  the  locus  of 
the  point  of  intersection,  P,  of  PjQ2  an(i  PzQi- 

3-  What  is  the  locus  of  P  in  the  preceding  example,  if  Qi 
and  Q2  are  always  on  opposite  sides  of  PiP2  ? 

4.  Do  Ex.  2,  §  2,  using  the  intercepts  of  the  moving  line  AB 
on  the  two  given  perpendicular  lines  as  auxiliary  variables. 

5.  The  same  for  Ex.  3,  §  2. 

6.  The  points  R  and  S  move,  one   on  each  of   two  fixed 
perpendicular  lines,  so  that  the  segment  RS  subtends  always 
a  right  angle  at  a  fixed .  point,  not  at  the  intersection  of  the 
two  lines.     Find  the  locus  of  the  mid-point  of  RS. 

Ans.  Perpendicular  bisector  of  the  line-segment  joining  the 
fixed  point  with  the  intersection  of  the  fixed  lines. 

7.  Two  right  angles,  having  their  vertices  in  fixed  points 
A  and  B,  rotate  about  these  points,  so  that  the  point  of  inter- 
section of   two  of   their  sides   traces  a  line   parallel   to  AB. 
What  is  the  locus  of  the  point  of  intersection  of  the  other  two 
sides  ? 

5.  Use  of  the  Formula  for  the  Sum  of  the  Roots  of  a  Quad- 
ratic Equation.  The  sum  of  the  roots  of  a  quadratic  equation 
(Ch.  IX,  §  5), 

Ax1  +  Ex  +  C  =  0,  A  =£  0, 

is  the  negative  of  the  ratio  of  the  coefficients  of  the  terms  in 
x  and  x1 ;  that  is, 

7J 

(1)  0?!  +  ^  =  --, 

where  xt  and  x2  are  the  roots. 

As  a  simple  example  of  the  way  in  which  this  fact  may  be 
used  to  advantage,  let  us  find  the  coordinates  of  the  point  P 
midway  between  the  points  of  intersection,  Px  and  P2,  of  a 
line  and  a  conic.  Take,  for  example,  the  line 

(2)  2x-y  =  l, 
and  the  ellipse 

(3)  3 


A  SECOND   CHAPTER  ON  LOCI 


275 


The  coordinates  (xly  y^)  and  (x2,  y2),  of  P:  and  P2,  are  the 
simultaneous  solutions  of  equations  (2)  and  (3).  Substituting 
in  (3)  the  value  of  y  from  (2)  and 
collecting  terms,  we  have  the  quad- 
ratic equation, 

19  x2- 


for  the  determination  of  xl  and  xz. 
We  are   interested,   not   in   the 
actual  values  of  xa  and  x.2,  but  in 
half    their   sum ;    for   this   is   the 
abscissa   of   the   mid-point,   P,   of 
PXP2.     By  (1)  the  sum  is  -^|.     Then  the  abscissa  of  P  is 
and,  since  P  lies  on  the  line  (2),  its  ordinate  is 


FIG.  6 


Consider  now  the  following  locus  problem  :  A  variable  tan- 

gent to  the  parabola, 
y~  =  2  mar, 
meets  the  hyperbola, 


in  the  points  PI  and  P2- 
What  is  the  locus  of  the  mid- 
point, P,  of  PiP-j  ? 

As  auxiliary  variables  we 
take  the  coordinates  (xf,  y')  of 
the  point  R  tracing  the  parab- 
ola  ;  they  satisfy  the  equa- 
tion of  the  parabola  : 


FIG.  7 


(4)  y'* 

The  tangent  at  R  has  the  equation 

(5)  y'y  =  m(x  +  a/). 

To  find  the  coordinates  of  P1  and  P2,  we  solve  (5)  simultane- 
ously with  the  equation  of  the  hyperbola.  Eliminating  y,  we 
have  : 

mx"  +  mx'x  —  c-y'  =  0. 


276  ANALYTIC   GEOMETRY 

Half  the  sum  of  the  roots  of   this   equation  is  X;  hence, 

^  (i), 

(6)  X==-|' 

Since  Y  is   the  ordinate  of  the  point  on  the  line  (5)  whose 
abscissa  is  given  by  (6),  we  have 

(7)  /r= 

We  now  have  three  equations,  (4),  (6),  and  (7),  from  which 
to  eliminate  the  auxiliary  variables  x'  and  y'.  We  solve  (6) 
and  (7)  for  x'  and  y',  obtaining 


Substituting  these  values  of  x'  and  y'  in  (4)  and  simplifying 

the  result,  we  have 

4F2=-raX. 

Consequently,  the  locus  of  P  is  a  parabola  with  vertex  at 
the  origin  and  opening  out  along  the  negative  axis  of  x  as 
axis.  The  origin  is  not  a  point  of  the  locus. 

EXERCISES 

1.  A  variable  tangent  to  the  circle 

a2  +  y2  =  a2 
meets  the  hyperbola 

2xy  =  a? 

in  the  points  Px  and  P2.     Find  the  equation  of  the  locus  of  the 
mid-point  of  PiP2.     Plot  the  locus. 

Ans.    —  |  —  =  —  ,  a  curve  which   does  not,  despite  its  ap- 
x2     y2     a2 

pearance,  consist  of  two  conjugate  rectangular  hyperbolas. 

2.  Two  equal  parabolas  have  the  same  axis  and  vertex,  but 
open  in  opposite  directions.     Find  the  locus  of  the  mid-points 
of  the  chords  of  one  which,  when  produced,  are  tangent  to  the 
other. 


A   SECOND   CHAPTER  ON  LOCI  277 

3.  Find  the  locus  of  the  mid-points  of  the  focal  chords — • 
chords  through  the  focus  —  of  a  parabola. 

Ans.  A  parabola  with  its  vertex  in  the  focus  of  the  given 
parabola,  with  the  same  axis,  but  half  the  size. 

4.  Determine  the  locus  of  the  mid-points  of  one  set  of  focal 
chords  of  an  ellipse. 

5.  The  same  for  a  hyperbola. 

6.  A  variable  tangent  to  a  parabola  meets  the  tangents  at 
the  extremities  of  the  latus  rectum  in  the  points  Pl  and  P2. 
Find  the  locus  of  the  mid-point  of  PiP2. 

7.  The  asymptotes  of  a  hyperbola   intercept  the  segment 
P\PZ  on  a   variable  tangent.     What  is  the  locus  of  the  mid- 
point of  P\P2  ?  Ans.   The  hyperbola  itself. 

6.  Loci  of  Inequalities.  Though  we  are  concerned  primarily 
in  mathematics  with  equalities,  it  is  not  infrequent  that  in- 
equalities become  important.  Accordingly,  it  is  not  out  of 
place  to  consider  here  the  loci  of  some  inequalities. 

Example  1.  The  equation  x  —  1  =  0  represents  all  the 
points  of  the  line  parallel  to  and  one  unit  to  the  right  of  the 
axis  of  y,  and  no  other  points.  Consequently,  the  inequality, 
x  —  1  =£  0,  represents  all  the  pojnts  of  the  plane  not  on  this 
line.  In  particular, 

x-l>0 

represents  all  the  points  to  the  right  of  it  and 

z-l<0 
represents  all  the  points  to  the  left  of  it. 

Example  2.  What  is  the  locus  of  points  whose  coordinates 
satisfy  the  inequality 

(1)  5«  +  12y  +  6>0? 

The  equation  obtained  by  replacing  the  sign  >  by  the  sign 
of  equality  represents  the  line  L  shown  in  the  figure.  From 


278  ANALYTIC   GEOMETRY 

Example  1  we  should  expect  that  the  locus  of  (1)  would  con- 
sist of  all  the  points  on  one  side  of  L.  This  is,  in  fact,  the 

case  :  If  the  quantity, 

F  =  5  x  +  12  y  -f  6, 

is  positive  for  a  certain  point 
(x0,  y0),  then  it  is  positive  for  all 
points  on  the  same  side  of  L  as 

(XQ,   V0). 

FIG    8 

We  prove  this  by  showing  that 

the  opposite  assumption  leads  to  a  contradiction.  Suppose 
that  F  becomes  negative  for  some  point  (x1}  y^)  on  the  same 
side  of  L  as  (x0)  y0).  Join  (o?0,  y0)  to  (x1}  yx)  by  any  curve  C 
not  cutting  L,  and  let  a  point  (#,  y)  trace  this  curve.  For 
(x0,  ?/0),  F  is  positive;  but  when  (x,  y)  has  reached  (xlt  y^),  F 
has  become  negative.  Consequently,  for  some  intermediate 
point  R  on  C,  F  has  the  value  zero,  inasmuch  as  its  value 
changes  continuously  as  (x,  y)  moves  along  C.  Hence  R  must 
lie  on  L,  —  a  contradiction,  since  we  took  C  as  a  curve  never 
cutting  L. 

To  ascertain  on  which  side  of  L  the  points  represented  by 

(1)  lie,  we  have  but  to  find  the  value  of  F  for  one  point  not 
on  L.     In  this  case  the  simplest  point  to  take  is  the  origin. 
But,  when  x  =  0  and  y  =  0,  Fjs  positive.     Therefore  the  locus 
of  (1)  consists  of  all  points  on  the  same  side  of  L  as  the  origin. 

Example  3.     What  is  the  locus  of  the  inequality 

(2)  y*  >2x? 

The  equation,  obtained  by  replacing  the  sign  >  by  the 
equality  sign,  represents  a  parabola.  By  the  reasoning  of 
Example  2,  then,  the  inequality  represents  all  the  points 
within,  or  all  the  points  without,  the  parabola.  The  latter  is 
clearly  the  case,  since  (2)  is  not  satisfied  by  the  coordinates 
of  the  point  (1,  0),  —  a  point  which  is  within  the  parabola. 


A   SECOND   CHAPTER  ON  LOCI 


279 


EXERCISES 

Find  the  loci  of  the  following  inequalities.  Draw  a  figure 
in  each  case  and  shade  the  area  of  points  represented  by  the 
inequality. 

1.   a  +  2>0.  2.   2?/  +  3<0. 

3.    x+  y  +  l  >  0.  4.   3z  —  4y  —  2  >  0. 

5.    2x  —  3y<Q.  6.   £2  +  y2<l. 

7.    f-  +  7 x  >  0.  8.   3 x2  +  4 0s  >  8. 

7.  Locus  of  Two  or  More  Simultaneous  Inequalities. 

Example  1.  Find  the  locus  of  points  whose  coordinates 
satisfy  simultaneously  the  two  inequalities 

(1)  5aj  +  12?/  +  6>0, 

(2)  3x-4y-2>0. 

Denote  the  left-hand  sides  of  (1)  and  (2)  by  FI  and  F2,  re- 
spectively. By  Example  2  of  §  6,  the  points  whose  coordi- 
nates satisfy  (1)  are  all  the 
points  which  are  on  the  same 
side  of  the  line  L^ :  FI  =  0  as 
the  origin  ;  similarly,  the  points 
whose  coordinates  satisfy  (2) 
are  all  the  points  which  are  on 
the  opposite  side  of  the  line 
L2  :  F2  =  0  from  the  origin. 
The  points  whose  coordinates  FIQ-  9 

satisfy    (1)    and    (2)    are    the 

points  common  to  these  two  sets,  namely,  those  of  region  I 
of  the  figure. 

Lying  between  the  lines  LY  and  L2  there  are  four  regions, 
I,  II,  III,  IV.  It  is  clear  from  the  foregoing  that  the  pairs  of 
simultaneous  inequalities  representing  these  regions  are : 

*   \    TF    -^   f\  •         *  *  1    TF     ^*  A  .  J.J-J-  •   i    -w-j  y-v  J.  v    .  s    _  ^. 

\  J?  9  ^  U  i  \  J?  o  <^  UI  lxfo<CUI  .To^lA 

\       &  ^  V*^  \       *    ^        "  \       *•  ^ 


280  ANALYTIC   GEOMETRY 

Example  2.  Find  the  points  satisfying  simultaneously  the 
inequalities  : 

?/2  —  2a;  <  0,         x  +  y  —  l<0. 

The  equations  obtained  by  replacing  the  signs  <  by  signs  of 
equality  represent  a  parabola  and  a  line  intersecting  it.  The 
locus  of  the  first  inequality  is  the  interior  of  the  parabola; 
that  of  the  second  is  the  half-plane  bounded  by  the  line  and 
containing  the  origin.  Common  to  these  two  regions  is  the 
finite  region  contained  between  the  parabola  and  the  line  ; 
this,  then,  is  the  locus  of  the  two  inequalities  taken  simul- 
taneously. 

EXERCISES 

Find  the  locus  of  points  whose  coordinates  satisfy  simul- 
taneously the  following  sets  of  inequalities.  Draw  a  figure  in 
each  case,  and  shade  the  region  represented. 


J4aj-3<0,  !5x-12y  +  26  >  0, 

\3x  +  2y-6  <  0.  |3z  +  4y-10>0. 

\2x-y  +  3  <0,  ja;2+?/2<4,       Ans. 

\4:X-2y+9  >  0.  }a?-3>0.    No  locus. 

|3a;2  +  4yJ-12  >  0,  |o;2-7y<0, 

5'     \2x-3y  +  12  >0.  ja;2-2/2  +  l>0. 

f2aj-y-3<0,  [^  +  ^-l>0, 

7.    |aj+3y-5<0,  8.     s  y  >  0, 

[5a;_j_y  +  3>  0.  [2x-y>0. 

Each  of  the  following  pairs  of  curves  divide  the  plane 
(minus  the  points  on  the  curves)  into  a  number  of  regions. 
Find  the  pairs  of  simultaneous  inequalities  representing  these 
regions. 

9.   5*/  +  8  =  0,  3x+8y-2=0. 

10.  5x-  12s/  +  26  =  0,        3x+4?/-10  =  0. 

11.  2z2  +  ?/2  =  8,  4:X-3y-2  =  Q. 

12.  '2  =  2mx  x  =  Q. 


A  SECOND   CHAPTER  ON  LOCI  281 

13.  x1  -y"-  =  4:,  2  a-  y  -2=0. 

14.  z-+4?/  =  0,  2x-3y-6  =  0. 

15.  ?/2  +  8o;  =  0,  z2+</2  =  9. 

8.   Bisectors  of  the  Angles  between  Two  Lines.     The  two  lines 


L2:  3x—    4y-2  =  0, 

are  given.     It  is  required  to  find  the  equations  of  the  lines  bi- 
secting the  angles  between  them. 

We  solve  this  problem  by  finding  the  locus  of  the  point 
P:(-X",  F)  moving  so  that  its  distance  DI  from  '  L\  equals  its 
distance  D2  from  Lz  : 

A  =  D2. 

According  to  Ch.  II,  §  8,  Z>x  and  D2  are 

n        ,  5.X  +  12F+6  n      ^3X-4F-2 

i/i   :=  T  -  ,  J_/n  :=  -\-  -  , 

13  5 

where,  in  'each  case,  that  sign  is  to  be  chosen  which  will  make 
the  distance  positive. 

The  lines  Lt  and  L2  are  those  of  Example  1,  §  7.  It  follows, 
from  the  results  there  given  in  connection  with  Fig.  9,  that 
the  signs  which  must  be  taken  to  make  DI  and  D2  both  positive 
are: 

If  P  is  in     I,  +  for  D1}  +  for  D2  ; 

if  P  is  in    II,  +  for  Dlt  —  for  D2  ; 

if  P  is  in  III,  —  for  DI,  —  for  D2  ; 

if  P  is  in  IV,  -  for  Dlt  +  for  D2, 

For,  if  P  lies,  for  example,  in  the  region  I,  then  the  numerators 
in  the  expressions  for  DI  and  D2  are  both  positive  and  the  + 
sign  must  be  taken  in  each  case  to  make  DI  and  D2  positive. 
If,  now,  P  is  in  I  or  III  and  Z>r  =  D2,  we  have 

5(5  X+  12  Y+  6)=  13(3  X  -  4  F-  2), 
or,  on  reducing, 
(1)  X-8F-4  =  0. 


282  ANALYTIC   GEOMETRY 

Thus  (1)  is  that  bisector  of  the  angles  between  Z/t  and  L2  which 
lies  in  the  regions  I  and  III. 

If  P  is  in  II  or  IV  and  Dt  =  Z)2>  we  have 

5(5  X  +  12  F  +  6)  =  -  13(3  X  -  4  F-  2) 
or 
(2)  16X  +  2F+1  =  0. 

This  is  the  bisector  which  lies  in  the  regions  II  and  IV. 

Simplification.  We  now  give  in  condensed  form  the  method 
of  finding  the  bisectors.  By  equating  DI  and  D2,  we  have 

5X+12Y+6=+3X-4:Y-2 
13  5 

If  we  take  both  signs  positive  or  both  negative  and  reduce  the 
result,  we  get  (1).  If  we  take  the  plus  sign  on  the  right  and 
the  minus  sign  on  the  left  or  vice  versa,  and  then  simplify,  we 
get  (2).  The  equations  (1)  and  (2)  represent  the  bisectors  ; 
which  equation  represents  a  chosen  bisector  is  easily  deter- 
mined by  making  a  plot. 

EXERCISES 

1.  Find  the  equations  of  the  bisectors  of  the  angles  between 
the  following  pairs  of  lines,  and  draw  a  figure  which  shall 
indicate  each  bisector. 


-  ,  -, 

\3x+    4^-10  =  0;  \    *-    0-4  =  0. 

2.  Find  the  equation  of  that  bisector  of  the  angle  between 
the  two  lines, 

4cc—  3?/  +  3  =  0        and         3x—  ±y  —  6  =  0, 

which  passes  through  the  region  between  the  two  lines  which 
contains  the  origin. 

3.  Find  the  equations  of  the  circles  tangent  to  the  lines  of 
Ex.  1,  Part  (a),  and  having  their  centers  on  the  line  y  =  8. 

4.  Find  the  equations  of  the  circles  tangent  to  the  lines  of 
Ex.  2  and  passing  through  the  point  (1,  0). 


A   SECOND   CHAPTER  ON  LOCI  283 

Given  the  triangle  ABC  with  the  sides 

AB:  3x  +  4y—    3  =  0, 

BC:  3x-±y-    3  =  0, 

CM:         12»-5y  +  15  =  0. 

5.  Prove  that  the  bisectors  of  the  interior  angles  of  the 
triangle  meet  in  a  point.     Find  its  coordinates.  Ans,  (—^,  0). 

6.  Find  the  equation  of  the  circle  iiascribed  in  the  triangle. 

Ans.     121(z2  +  f)  _j_  88  x  -  65  =  0. 

7.  Show  that  the  bisector  of  the  interior  angle  at  the  vertex 
A  and  the  bisectors  of  the  exterior  angles  at  the  vertices  B  and 
C  meet  in  a  point.     Find  its  coordinates.  Ans.     (1,  —  5). 

8.  Find  the  equation  of  the  circle  tangent  to  BC,  and  to  AB 
and  AC  produced.  Ans.     x1  +  y-  —  2  x  +  10  y  +  10  =  0. 

9.  How  many  circles  are  there  tangent   to   three   lines? 
Draw  a  figure  showing  these  circles. 

10.  Given  the  triangle  with  vertices  A,  B,  and  C  in  the  three 
points  (1,  0),  (—2,  4),  and  (—  5,  —  8).  Prove  analytically  that 
the  bisector  of  the  interior  angle  at  A  divides  the  side  BC  into 
segments  proportional  to  AB  and  AC. 

See  also  Exs.  26-30  at  the  end  of  the  chapter. 

'EXERCISES  ON  CHAPTER  xm 

1.  Let  AA'  be  a  fixed  diameter  of  a  circle  and  R  a  point 
tracing  the  circle.     Find  the  locus  of  the  point  of  intersection 
of  A'R  and  the  line  through  A  perpendicular  to  the  tangent  at 
E. 

2.  Find  the  locus  of  the  point  of  intersection  of  the  nor- 
mals to  an  ellipse  and  to  the  auxiliary  circle  at  corresponding 
points.     Take  the  eccentric  angle  (Ch.  VII,  §  10)  as  the  aux- 
iliary variable. 

3.  The  circle  xz  +  y*  =  a2  cuts  the  axis  of  y  in  A :  (0,  a). 
A  point  S  traces  the  tangent  at  A  and  the  second  tangent  from 
S  touches  the  circle  in  R.     Find  the  locus  of  the  point  of  in- 
tersection of  the  altitudes  of  the  triangle  ARS. 


284  ANALYTIC   GEOMETRY 

Suggestion.  Take  the  abscissa  of  S  and  the  coordinates  of 
R  as  auxiliary  variables,  and  use  the  fact  that  OS  is  perpen- 
dicular to  AR. 

4.  Let  AA'  be  a  fixed  diameter  of  a  circle  and  R  a  point 
tracing  the  circle.     Find  the  locus  of  the  point  of  intersection 
of  AR  and  the  line  joining  A'  to  the  point  of  intersection  of 
the  tangents  at  A  and  R. 

5.  The  normal  to  a  hyperbola  at  a  variable  point  R  meets 
the  transverse  axis  in  N.    Determine  the  locus  of  the  mid- 
point of  RN, 

6.  Find  the  locus  of  the  point  of  intersection  of  the  line 
drawn  through  one  focus   of  an  ellipse  perpendicular  to  a 
variable   tangent  and   the  line  drawn  through  the  point  of 
tangency  parallel  to  the  transverse  axis. 

Ans.  An  ellipse,  center  in  the  focus  chosen,  with  axes 
having  the  same  directions  as  those  of  the  given  ellipse. 

7.  Find  the  locus  of  the  point  of  intersection  of  the  line 
drawn  through  one  vertex  of  a  hyperbola  perpendicular  to  a 
variable  tangent  and  the  line  drawn  through  the  point  of  tan- 
gency parallel  to  the  transverse  axis. 

8.  Find  the  locus  of  the  point  of  intersection  of  the  line 
drawn  through  the  focus  of  a   parabola   perpendicular   to  a 
variable  tangent  and   the   line   joining   the  vertex  with   the 
point  of  tangency. 

Ans.  An  ellipse,  whose  minor  axis  is  the  line-segment  join- 
ing the  vertex. of  the  parabola  to  the  focus. 

9.  Two  lines,  passing  through  the  points  A  and  B  respec- 
tively,  are   originally   in   coincidence   along   AB.     They  are 
made  to  rotate  in  the  same  direction  about  A  and  B  respec- 
tively, the  first  twice  as  fast  as  the  second.     What  is  the  locus 
of  their  point  of  intersection  ? 

10.  Find  the  locus  of  the  center  of  a  circle  which  touches 
one  of  two  perpendicular  lines  and  intercepts  a  segment  of 
constant  length  on  the  other. 


A   SECOND    CHAPTER   ON    LOCI  285 

11.  Find  the  locus  of  the  point  of  intersection  of  the  line 
drawn  through  the  vertex  of  the  parabola  yz  =  2  rnx  perpen- 
dicular to  a  variable  tangent  and  the  line  drawn  through  the 
point  of  tangency  perpendicular  to  the  axis. 

Ans.    The  semi-cubical  parabola,  my2=  2  a?. 

12.  A  vertex  0  of  a  quadrilateral  and  the  directions  of  the 
sides  through  0  are  fixed.     The  two  angles  adjacent  to  0  are 
right  angles  and  the  diagonal  joining  their  vertices  has  a  fixed 
direction.     Find  the  locus  of  the  fourth  vertex. 

Ans.  Straight  line  through  0,  perpendicular  to  the  line 
through  0  which  makes  an  angle  with  the  fixed  direction  equal 
to  the  sum  of  the  two  angles  which  the  sides  through  0  make 
with  the  fixed  direction. 

13.  A  parallelogram  has  sides  of  constant  length  a  and  b 
and  has  one  vertex  fixed  at  a  point  0.     It  opens  and  closes 
so  that  the  two  sides  through  0  are  always  equally  inclined 
to  a  fixed  line  throiigh  0.     Taking  the  angle  which  these  sides 
make  with  the  fixed  line  as  auxiliary  variable,  find  the  locus 
of  the  vertex  opposite  to  0. 

14.  Each  of  two  straight   lines   moves  always    parallel  to 
itself  so  that  the  product  of  the  distances  of  the  lines  from  a 
fixed  point  0  is  constant.     Find  the  locus  of  their  point  of 
intersection,  taking  the  axes  so  that  0  is  the  origin  and  the 
directions  of  the  two  lines  are  equally  inclined  to  the  axis  of  x. 

Ans.  Two  conjugate  hyperbolas,  center  at  0,  with  asym- 
ptotes parallel  to  the  fixed  directions. 

15.  Find  the  locus  of  the  center  of  a  circle  which  passes 
through  a  fixed  point  on  one  of  two  perpendicular  lines  and 
intercepts  a  segment  of  constant  length  on  the  other. 

16.  Find  the  locus  of  points  from  which  it  is  possible  to 
draw  two  perpendicular  normals  to  a  parabola. 

17.  Find  the  locus  of  the  point  of  intersection  of  the  tangents 
to  an  ellipse  at  points  subtending  a  right  angle  at  the  center. 

18.  The  preceding  problem  for  the  hyperbola. 


286  ANALYTIC   GEOMETRY 

19.  Determine   the   locus  of   the   mid-point  of  a   variable 
chord  of  an  ellipse  drawn  from  a  vertex. 

20.  Find  the  locus  of  the  mid-point  of  a  variable  chord  of 
a  parabola  which  subtends  a  right  angle  at  the  vertex. 

21.  A  point  R  traces  an  ellipse,  of  which  A  and  A'  are  the 
vertices.     Find  the  locus  of  the  point  of  intersection  of  the 
lines  drawn  through  A  and  A  perpendicular  to  AR  and  A'R 
respectively. 

Ans.  An  ellipse,  similar  to  and  with  the  same  center  as  the 
given  ellipse,  but  with  opposite  transverse  and  conjugate  axes. 

22.  The  asymptotes  of  a  hyperbola  intercept  the  segment 
AB  on  a  variable  tangent.     What  is  the  locus  of   the  point 
dividing  AB  in  a  given  ratio,  w^  :  ra2  ? 

Ans.  A  similar  hyperbola,  with  the  same  transverse  and 
conjugate  axes. 

Exercises  23-25.  Determine  the  equations  of  the  desired  loci 
by  use  of  rectangular  coordinates.  To  identify  the  locus  from 
its  equation  introduce  polar  coordinates. 

23.  Find  the  locus  of  the  point  of  intersection  of  a  variable 
tangent  to  a  rectangular  hyperbola  with  the  line  through  the 
center  perpendicular  to  the  tangent.  Ans.   A  lemniscate. 

24.  Find  the  locus  of  the  point  of  intersection  of  a  variable 
tangent  to  the  circle  x1  +  y1  +  2  ax  =  0  and  the  perpendicular 
to  this  tangent  from  the  origin.  Ans.   A  cardioid. 

25.  What  is  the  locus  of  the  mid-points  of  the  chords  of  the 
circle  x1  -f  y1  =  a2  which,  when  produced,  are  tangent  to  the 
hyperbola  2  xy  =  c-  ?  Ans.    A  lemniscate. 

26.  Show  that  the  line 

x  cos  30°  +  y  sin  30°  =  5 

is  5  units  distant  from  the  origin  and  that  the  perpendicular 
from  the  origin  to  it  makes  with  the  positive  axis  of  x  an  angle 
of  30°.  Prove  that  the  distance  of  the  point  (XQ,  y0)  from  the 

line  is 

-  (XQ  cos  30°  -f  y0  sin  30°  -  5), 


A  SECOND  CHAPTER  ON  LOCI  287 

if  (x0,  y0~)  is  on  the  same  side  of  the  line  as  the  origin,  and  is 

XQ  cos  30°  +  2/0  sin  30°  -  5, 
if  (x0,  y0)  is  on  the  opposite  side  of  the  line  from  the  origin. 

27.  State  and  prove  for  the  line 

(1)  xcos<f>  +y  sin<£  =  p,  p>Q 

the  results  corresponding  to  those  given  in  the  preceding  exer- 
cise for  the  particular  line  for  which  <f>  —  306,  p  =•  5.  Prove 
that  the  equation  of  every  line  can  be  written  in  the  form  (1). 

28.  Two  lines,  with  their  equations  in  the  form  (1),  are 
given.    Let  a  =  0,  /3  =  0  be  the  abridged  notation  (Ch.  IX,  §  3) 
for  these  equations.     Prove  that  the  bisectors  of  the  angles 
between  the  two  lines  are  given  by  the  equations  a  —  ft  =  0 
and  a  +  ft  =  0.     Show  that,  if  neither  line  goes  through  the 
origin,  the  bisector  a  —  ft  =  0  passes   through   that   opening 
between  the  lines  in  which  the  origin  lies, 

29.  The  equations  of  the  sides  of  a  triangle,  given  in  the 
form  (1),  are  a  =  0,  /3  =  0,   and  y  =  0.      Assuming   that  the 
origin  lies  within  the  triangle,  find  the  equations  of  the  bisec- 
tors of  the  interior  angles  and  prove  that  they  meet  in  a  point. 

30.  Prove  that  the  bisectors  of  two  exterior  angles  of  the 
triangle   of  the   preceding   exercise   and  the  bisector  of  the 
interior  angle  at  the  third  vertex  meet  in  a  point. 


CHAPTER   XIV 


DIAMETERS.     POLES   AND   POLARS 

1.  Diameters  of  an  Ellipse.  By  the  axes  of  an  ellipse  we 
may  mean  either  the  transverse  and  conjugate  axes,  indefinite 
straight  lines,  or  the  major  and  minor  axes,  the  segments  of 
these  lines  intercepted  by  the  ellipse ;  cf.  the  dual  definition, 
Ch.  VII,  §  1. 

By  a  diameter  of  an  ellipse  we  may  mean,  also,  one  of  two 
things,  either  an  indefinite  straight  line  through  the  center  of  the 
ellipse,  or  the  segment  of  this  line  intercepted  by  the  ellipse ;  and 
we  agree  to  adopt  this  dual  definition.  The  length  of  the  seg- 
ment is  called  the  length  of  the  diameter;  its  end  points,  the 
extremities  of  the  diameter. 

Problem.  What  is  the  locus  of  the  mid-points  of  a  set  of 
parallel  chords  of  an  ellipse  ? 

In  the  special  case  of  a  circle,  the  locus  is  a  diameter,  con- 
sidered as  a  line-segment.  This  is  true,  also,  for  the  general 
ellipse.  For,  if  the  chords  are  parallel  to  an  axis  of  the 
ellipse,  the  theorem  is  geometrically 
obvious ;  if  they  are  oblique  to  the 
axes,  as  is  generally  the  case,  we 
resort  to  an  analytical  proof. 
Let  the  ellipse  be 


FIG.  1 


(1) 


and  let  A.  (=£  0)  be  the  slope  of  the  chords.     Consider  a  variable 
chord  of  slope  A.  moving  always  parallel  to  itself.      Its 

288 


mo- 


DIAMETERS.     POLES  AND   POLARS  289 

tion  we  express  analytically  by  taking  ft,  its  intercept  on  the 
axis  of  y,  as  auxiliary  variable.  The  equation  of  the  chord  is, 
then, 

where  A  is  constant  and  ft  is  variable.  » 

The  work  now  proceeds  according  to  the  method  of 
Ch.  XIII,  §  5.  If  in  (1)  we  set  for  y  its  value  as  given  by 
(2),  we  obtain  the  equation 

or  (a2A2  +  &2)»2  +  2  a2  A/to  +  a\^  —  ft2)  =  0, 

whose  roots  are  the  abscissae  of  the  two  points  of  intersection 
of  the  line  (2)  with  the  ellipse.  Half  the  sum  of  these  roots 
is  X,  the  abscissa  of  the  mid-point,  P,  of  the  chord.  Hence, 
by  the  formula,  Ch.  XIII,  §  5,  (1),  for  the  sum  of  the  roots  of 
a  quadratic  equation, 

(3)  X  = *M_. 

a2A2  +  ft2 

Since,  moreover,  P:  (X,  F)  lies  on  the  chord  (2),  we  have 

It  remains  to  eliminate  ft  from  (3)  and  (4).  Substituting 
its  value,  as  given  by  (4),  into  (3)  and  simplifying  the  result- 
ing equation,  we  obtain 

(5)  62x+a2AF=0. 

This  is  the  equation  of  a  line  through  the  center  of  the 
ellipse,  that  is,  a  diameter.  It  is  clear  geometrically,  however, 
that  it  is  not  the  indefinite  line  which  is  the  locus,  but  merely 
the  portion  of  it  lying  within  the  ellipse.  We  have  thus  ob- 
tained the  following  result. 

THEOREM  1.  The  locus  of  the  mid-points  of  a  set  of  parallel 
chords  of  the  ellipse  (1)  is  a  diameter,  considered  as  a  line-seg- 
ment (exclusive  of  the  end  points').  If  the  slope  of  the  chords  is 
A(=£  0),  the  slope  A'  of  the  diameter  is 

(6)  A'  =  -^r- 


290  ANALYTIC   GEOMETRY 

EXERCISES 

1.   Find  the  locus  of  the  mid-points  of  the  chords  of  the 
ellipse 


which  are  inclined  at  an  angle  of  135°  to  the  axis  of  x.  First 
draw  an  accurate  figure,  showing  the  chords  and  the  locus  ;  then 
solve  the  problem  analytically,  using  the  method,  but  not  the 
formulas,  of  the  text. 

2.  Prove  the  converse  of   Theorem  1,  namely,   that  every 
diameter  of  the  ellipse  (1)  bisects  some  set  of  parallel  chords. 
Show  that,  if  A'(=£  0)  is  the  slope  of  the  diameter,  then  the 
chords  which  it  bisects  are  of  slope  A,  where 

\=      -*-. 
a2A' 

3.  Prove  analytically  that  the  tangent  to  an  ellipse  at  an 
extremity  of  a  diameter  is  parallel  to  the  chords  which  the 
diameter  bisects. 

Suggestion.  Let  (x1}  y:~)  be  the  coordinates  of  the  extremity 
of  the  diameter  and  find,  by  using  (6),  the  slope  A  of  the 
chords  in  terms  of  a^  and  y{. 

2.  Conjugate  Diameters  of  an  Ellipse.  Two  mutually  per- 
pendicular diameters  of  a  circle  have  the  property  that  each 
bisects  the  chords  parallel  to  the  other.  The  axes  of  an  el- 
lipse have  this  same  property.  Are  there  other  pairs  of 
diameters  of  the  ellipse  which  have  it?  This  question  is 
answered  in  the  affirmative  by  the  following  theorem. 

THEOREM  2.     If  one  diameter  bisects  the 
chords  parallel  to  a  second,  the  second  di- 
ameter bisects  the  chords  parallel  to  the  first. 
The  two  diameters  stand  in  a  reciprocal 
relationship  ;  each  bisects  the  chords  paral- 
lel to  the  other.     We  call  them  a  pair  of 
conjugate   diameters,  and    say   that   each   is   conjugate   to   the 
other. 


DIAMETERS.     POLES  AND   POLARS  291 

We  now  prove  Theorem  2.  Let  the  diameter  D'  bisect  the 
chords  parallel  to  the  diameter  D  ;  to  prove  that  D  bisects 
the  chords  parallel  to  D'. 

Denote  the  slopes  of  D  and  D'  by  A  and  A'.  By  hypothesis, 
the  diameter  of  slope  A'  bisects  the  chords  of  slope  A  ;  conse- 
quently, by  Th.  1,  §  1, 

A'  —  -*- 

a2A 

But  then  x  =  —  —  • 

a2A' 

This  equation  says  that  the  diameter  of  slope  A  bisects  the 
chords  of  slope  A'  ;  that  is,  D  bisects  the  chords  parallel  to  Z)', 
q.  e.  d. 

Incidentally,  we  have  also  proved  the  following  theorem. 

THEOREM  3.     Two  diameters  D  and  D'  of  the  ellipse 


are  conjugate,  if  and  only  if  they  are  the  axes  or  have  slopes  A 
and  A'  related  by  the  equation 

(2)  AA'=-£. 

az 

The  symmetry  of  (2)  in  A  and  A'  corresponds  to  the  sym- 
metry in  the  geometrical  relationship  of  D  and  D'. 

To  each  diameter  D  there  corresponds  a  conjugate  diameter 
—  the  diameter  parallel  to  the  chords  which  D  bisects.  There 
are,  then,  infinitely  many  pairs  of  conjugate  diameters.  Since, 
by  (2),  the  product  of  the  slopes  of  any  pair,  other  than  the 
axes,  is  negative,  the  two  diameters  of  the  pair  pass  through 
different  quadrants. 

The  axes  are  the  only  mutually  perpendicular  pair,  unless 
the  ellipse  becomes  a  circle.  For,  if  any  other  pair  were  per- 
pendicular, the  product,  AA',  of  their  slopes  would  be  —  1,  and 
this  is  impossible,  according  to  (2),  unless  62  =  a2  ;  but  then 
the  ellipse  becomes  a  circle. 


292 


ANALYTIC  GEOMETRY 


We  now  find  the  conjugate  diameters  which  are  equally 
inclined  to  the  axes.  It  is  evident,  geometrically,  that  these 
will  also  be  the  conjugate  diameters  of  equal  lengths.  If  they 
exist,  their  slopes  must  be  equal  except  for  sign :  X'  =  —  X. 
Hence,  by  (2), 

.  „      52  j 

X2  =  —  and  X  =  ±  — 

a2  a 

We  see,  then,  that  there  is  a  single  pair  of  conjugate  diameters 
which  are  equally  inclined  to  the  axes,  or  have  equal  lengths. 
They  are  the  diagonals  of  the  rectangle  circumscribed  about 

tKe  ellipse  (Fig.  3).     We  denote 
them  by  DI  and  Z)/. 

Now  let  a  diameter  D,  starting 
from  coincidence  with  the  trans- 
verse axis  AA',  rotate  about  0 
into  coincidence  with  DI  ;  then 
the  conjugate  diameter,  D',  starts 
from  coincidence  with  the  con- 
jugate axis  BB',  and  rotates  into 

coincidence  with  Z)/.  But  D'  rotates  more  quickly  than  D,* 
so  that  the  angle  from  D  to  D',  at  first  90°,  becomes  obtuse 
and  steadily  increases.  When  D  continues  to  rotate  from  DI 
to  BB',  then  D'  rotates  from  Z>/  to  AA' ;  but  now  D'  rotates 
less  quickly  than  Z>,  so  that  the  angle  from  Z>  to  D'  decreases 
and  becomes  again  90°  in  the  final  position. 

EXERCISES 

1.  Draw  accurately  an  ellipse  whose  axes  are  10  cm.  and 
7  cm.  Construct  the  axes  and  the  pair  of  conjugate  diameters' 
equally  inclined  to  the  axes.  Then  draw  the  diameters  in- 


*  Since  the  slope  of 


is  -  <  1,  X  A'ODi  <  45°  and  X  B'ODi'  >  45° 

a 


hence  D  has  a  smaller  angle  through  which  to  rotate  than  D'  .  Con- 
sequently, it  is  to  be  expected  that  D'  will  rotate  more  quickly  than  D. 
A  proof  of  the  fact  may  easily  be  given  later,  when  the  student  studies 
the  Calculus. 


DIAMETERS.     POLES  AND   POLARS  -  293 

clined  at  angles  of  10°,  20°,  30°,  40°,  50°,  60°,  70°,  and  80°'  to  the 
transverse  axis.  For  each  of  these  diameters  compute  the 
slope,  and  the  angle  of  inclination,  of  the  conjugate  diameter 
and  construct  it.  Find  the  angles  between  the  successive 
diameters  of  this  new  set,  and  also  the  angles  between  the 
successive  pairs  of  conjugate  diameters.  Mark  clearly  the 
pairs,  and  study  the  results  and  the  figure  in  light  of  the  text. 

2.  Prove  that,  if  one  of  a  pair  of  conjugate  diameters  of 
an  ellipse  has  the  slope  e  or  —  e,  where  e  is  the  eccentricity, 
the  other  joins  two  extremities  of  the  latera  recta. 

3.  If  the  equal  conjugate  diameters  of  an  ellipse  form  with 
one  another  an  angle  of  60°,  what  is  the  eccentricity  of  the 
ellipse  ? 

4.  The  axes  of  an  ellipse  are  the  axes  of  coordinates  and 
the  slopes  of  two  conjugate  diameters  are  -|  and  —  |.     What  is 
the  eccentricity  ? 

5.  The  same,  if  the  slopes  of  two  conjugate  diameters  are 
—  f  and  f . 

6.  Prove  that  the  line  joining  a  focus  to  the  point  of  inter- 
section of  the  corresponding  directrix  and  a  diameter  is  per- 
pendicular to  the  conjugate  diameter. 

3.  Diameters  of  a  Hyperbola.  A  diameter  of  a  hyperbola  is 
denned  in  the  same  way  as  a  diameter  of  an  ellipse,  §  1. 
Certain  diameters  of  a  hyperbola,  however,  do  not  meet  the 
curve.  Special  definitions  of  the  length  and  extremities  of  such 
a  diameter  must,  then,  be  adopted.  These  we  shall  consider 
later. 

The  locus  of  the  mid-points  of  a  set  of  parallel  chords  of 
slope  A.  (=£  0)  of  the  hyperbola 

(1)  i*_^=l 

a2     62 

can  be  found  by  the  method  of  §  1.  It  is,  however,  unneces- 
sary to  repeat  the  work  there  given.  For,  this  work  becomes 


294 


ANALYTIC   GEOMETRY 


valid  immediately  for  the  hyperbola  (1)  if,  in  it,  we  replace 
62  by  —  62.     It  follows,  then,  that  the  locus  now  required  is 
the  diameter 
(2)  -  Ifa  +  a?Xy  =  0         or         62cc  -  a2Ay  =  0. 

The  locus  consists  of  all  the  points  of  this  diameter  only  if 
the  given  chords  connect  points  of  opposite  branches  of  the 


FIG.  4 


FIG.  5 


hyperbola  (Fig.  4).  If  the  chords  connect  points  on  the  same 
branch  (Fig.  5),  the  locus  is  merely  the  points  of  the  diameter 
which  lie  within  the  curve.  The  result  can  be  stated  as  follows. 
THEOREM  4.  Tlie  locus  of  the  mid-points  of  a  set  of  parallel 
chords  of  the  hyperbola  (1)  is  a  diameter,  or  so  much  of  a  diame- 
ter as  lies  ivithin  the  curve.  If  the  slope  of  the  chords  is  A  ( =£  0), 
the  slope  A'  of  the  diameter  is 

(3)  x'=^-          -.T'1 

There  are  chords  of  an  ellipse  with  any  given  direction. 
This  is  not  true,  however,  for  a  hyperbola.  For,  there  are  no 
chords  of  a  hyperbola  parallel  to  an  asymptote,  since  a  line 
parallel  to  an  asymptote  meets  the  curve  in  but  one  point. 
Consequently,  the  slope,  A,  of  the  chords  of  Theorem  4  cannot 
have  either  of  the  values,  ±  b/a. 

EXERCISES 
1.    A  set  of  parallel  chords  of  the  rectangular  hyperbola 

x2  -  y*-  =  6 

are  inclined  at  an  angle  of  30°   to  the   positive   axis  of  x. 
What  is  the  inclination  of  the  diameter  which  bisects  them  ? 


DIAMETERS.     POLES  AND   POLARS  295 

First  draw  an  accurate  figure,  showing  the  chords  and  the 
diameter ;  then  solve  the  problem  analytically,  without  refer- 
ence to  the  formulas  of  the  text. 

2.  If  a  set  of  parallel  chords  has  a  slope  nearly  equal  to 
that  of  an  asymptote  S,  then  the  diameter  D  bisecting  the 
chords  has  a  slope  nearly  equal  to  that  of  S,  and  when  the 
chords  approach  a  limiting  position  of  parallelism  to  S,  then 
D  approaches  S  as  its  limit.     Draw  a  figure  showing  the  rea- 
sonableness of  this  theorem  and  then  prove  the  theorem  ana- 
lytically by  use  of  (3).  \ 

3.  Prove    the    converse    of    Theorem    4,   namely :    Every 
diameter  of  a  hyperbola,  not  an  asymptote,  bisects  some  set 
of  parallel  chords.     Cf.  §  1,  Ex.  2. 

4.  Show  that  the  mid-point  of  a  chord  of  a  hyperbola  is  also 
the  mid-point  of  the  chord  of  the  conjugate  hyperbola  which 
lies  on  the  same  line.     Hence  show  that  the  mid-points  of  the 
chords  of  a  given  slope  lie  on  one  and  the  same  diameter, 
whether  the  chords  are  chords  of  the  given  hyperbola  or  of 
its  conjugate. 

5.  Prove  Ex.  3,  §  1,  for  a  hyperbola. 

4.  Conjugate  Diameters  of  a  Hyperbola.  Let  the  diameter, 
ZX,  of  slope  A',  bisect  the  chords  of  the  hyperbola 

(1)  *_£=! 

a2     62 

which  are  parallel  to  the  diameter  D,  of  slope  A(=£  0).  Then, 
by  Th.  4,  §  3, 

X'-*-  or  XV.*. 

a2A  a2 

Since  these  equations  are  symmetric  in  A  and  A',  it  follows  that 
the  diameter  D  bisects  the  chords  parallel  to  D'. 

Thus  Theorem  2,  §  2,  is  established  for  the  hyperbola,  and 
the  two  diameters  D  and  D'  are,  in  the  sense  of  that  theorem, 
conjugate  diameters;  each  bisects  the  chords  parallel  to  the 
other. 


296  ANALYTIC   GEOMETRY 

We  have  also  proved,  incidentally,  the  following  theorem. 

THEOREM  5.  Two  diameters,  D  and  D',  of  the  hyperbola  (1) 
are  conjugate  if  and  only  if  they  are  the  axes  or  have  slopes,  \ 
and  A',  related  by  the  equation 

(2)  AV=|. 

There  are  infinitely  many  pairs  of  conjugate  diameters,  as 
in  the  case  of  the  ellipse.  But  here  the  two  diameters  of  a 
pair,  not  the  axes,  pass  through  the  same  quadrants,  since  the 
product,  XA',  of  their  slopes  is  positive. 

The  value,  62/a2,  of  this  product  is  the  square  of  the  slope 
of  an  asymptote.     The  slope  of  an  asymptote,  therefore,  is  a 
mean  proportional  between  the  slopes 
of  any  two  conjugate  diameters,  not 
the    axes.      Consequently,    two    such 
conjugate   diameters,   D  and   D',  are 
always  separated  by  the  asymptote  S 
which  lies  in  the  same  quadrants  with 
Fl'G  6  them,  and  the  nearer  D  lies  to  S  on 

the  one  side,  the  nearer  D'  will  lie  to 

S  on  the  other  side.  If  D  approaches  S  as  a  limiting  position, 
then  so  will  D'.  Thus,  an  asymptote  is  often  spoken  of  as  a 
self-conjugate  diameter  ;  actually,  however,  it  has  no  conju- 
gate, since,  as  we  have  seen,  there  are  no  chords  parallel  to  it. 
It  is  now  clear  that  if  a  diameter,  D,  starting  from  coinci- 
dence with  the  transverse  axis,  rotates  in  one  direction  about 
0  into  coincidence  with  an  asymptote,  then  the  conjugate 
diameter,  D',  starting  from  the  conjugate  axis,  will  rotate  in 
the  opposite  direction  about  0  into  coincidence  with  the  same 
asymptote. 

Conjugate  Hyperbolas.     Consider  now  the  hyperbola, 


conjugate   to   the   hyperbola  (1).     Since  the  two   hyperbolas 
have  the  same  center,  they  have  the  same   diameters,  con- 


DIAMETERS.     POLES  AND   POLARS 


297 


sidered  as  indefinite  straight  lines.  Moreover,  they  have  the 
same  pairs  'of  conjugate  diameters.  For,  the  chords  of  (3)  of  a 
given  slope  and  the  chords  of  (1)  of  the  same  slope  are  bisected 
by  one  and  the  same  diameter,  Ex.  4,  §  3. 

We  see  now  a  suitable  definition  for  the  extremities  of  a 
diameter  which  does  not  meet  the  given  hyperbola.  They 
shall  be  the  points  in  which  the  diameter  meets  the  conjugate 
hyperbola  (Fig.  7),  and  the  distance  between  these  points  shall 
be  the  length  of  the  diameter.* 

Conjugate  Diameters  of  a  Rectangular  Hyperbola.  A  special 
ellipse,  all  of  whose  conjugate  diameters  are  mutually  perpen- 
dicular, is  the  circle.  There  is  no  special 
hyperbola  with  this  property,  since  two 
conjugate  diameters  of  a  hyperbola,  other 
than  the  axes,  always  pass  through  the 
same  quadrants.  For  this  reason,  too, 
there  are  no  conjugate  diameters  equally 
inclined  to  the  axes. 

There  may,  however,  be  conjugate  di- 
ameters, each  of  which  has  the  same 
inclination  to  one  axis  as  the  other  has  to  the  other  axis.  The 
product,  XA',  of  the  slopes  of  two  such  diameters  is  1 ;  hence, 
by  (2),  such  diameters  exist  only  if 


FIG.  7 


or 


a 


a2  =  62, 


that  is,  only  if  the  hyperbola  is  rectangular.  In  this  case 
AA'  =  1,  and  every  pair  of  conjugate  diameters  are  in  the  re- 
quired relation.  Consequently,  the  two  diameters  are  equally 
inclined  to  the  asymptotes,  inasmuch  as  the  asymptotes  are 
now  the  bisectors  of  the  angles  between  the  axes.  They  are 
also  equal  in  length,  as  considerations  of  symmetry  immedi- 
ately show  (Fig.  7).  We  have  thus  proved  the  following 
theorem. 


*  An  asymptote,  considered  as  a  diameter,  we  shall  not  think  of  as 
having  length  or  extremities.  • 


298  ANALYTIC  GEOMETRY 

i 

THEOREM  6.  Two  conjugate  diameters  of  a  rectangular  hyper- 
bola are  always  equally  inclined  to  the  asymptotes  and  always 
equal  in  length. 

It  can  be  shown  that  the  rectangular  hyperbola  is  the  only 
one  with  either  of  these  properties.  Cf .  Ex.  6,  below,  and  §  6, 
Ex.  5.  Thus  the  rectangular  hyperbola  plays  a  role  among  the 
hyperbolas  which  is  somewhat  similar  to  that  played  by  the 
circle  among  the  ellipses. 

EXERCISES 

1.  Draw  accurately  the  hyperbola  for  which  2  a  =  10  cm.  and 
26  =  7  cm.     Construct  the  axes,  AA'  and  BB',  the  asymptote 
S  passing  through  the  first  quadrant,  and  the  diameters  Z>15  D2, 
Ds  inclined  at  angles  of  10°,  20°,  30°  to  the  transverse  axis. 
Compute  the  slopes  and  angles  of  inclination  of  the  conjugate 
diameters,  £)/,  D2',  D3',  and  draw  these  diameters.     Find  the 
angles  between  the  successive  diameters,  BB',  Z)/,  D2',  D3',  S, 
and  compare  them  with  the  corresponding  angles  between  the 
diameters,  AA',  D1}  D.2,  D3,  S.     Study  the  results  and  the  fig- 
ure in  light  of  the  text. 

2.  Prove  Ex.  2,  §  2,  for  the  hyperbola. 

3.  Prove  that  the  asymptotes  of  the  hyperbola 

^2_2/2_ 

a2     b*~ 

are  conjugate  diameters  of  the  ellipse 

*4-£  =  1 

a2     &2 

4.  The  axes  of  a  hyperbola  are  the  axes  of  coordinates,  and 
the  slopes  of  two  conjugate  diameters  are  2  and  f .     What  is 
the  eccentricity  of  the  hyperbola?     Two  answers. 

5.  Prove  Ex.  6,  §  2,  for  the  hyperbola. 

6.  Show  that  two  conjugate  diameters  of  a  hyperbola  are 
never  equally  inclined  to  the  asymptotes  unless  the  hyperbola 
is  rectangular. 


DIAMETERS.     POLES  AND   POLARS 


299 


I  I  I  I  I II 


5.  Diameters  of  a  Parabola.  When  one  focus  and  the  cor- 
responding directrix  of  a  central  conic  —  an  ellipse  or  a  hyper- 
bola—  are  held  fast  and  the  center  is  allowed  to  recede  in- 
definitely along  the  transverse  axis,  the  limit  of  the  cdnic  is  a 
parabola  and  the  limit  of  the  diameters  of  the  conic  is  a  set 
of  lines  parallel  to  the  axis  of  the  parabola. 
Accordingly,  by  a  diameter  of  a  parabola  we 
shall  mean  any  line  in  the  direction  of  the  axis 
of  the  parabola. 

If  this  definition  is  really  in  accord  with 
that  of  a  diameter  of  a  central  conic,  we 
should  find  that  the  mid-points  of  a  set  of 
parallel  chords  of  a  parabola  lie  on  a  line  in 
the  direction  of  the  axis.  This  is  the  case. 
If  the  chords  are  perpendicular  to  the  axis, 
their  mid-points  evidently  lie  on  the  axis ;  if  the  slope  of  the 
chords  is  A(=£  0),  and  the  equation  of  the  parabola  is 

(1)  y1  =  2  mx, 

the  mid-points  of  the  chords  lie  on  the  line 

(2)  2/  =  ™, 

as  may  easily  be  shown.  4 

EXERCISES 

1.  Establish  the  result  embodied  in  formula  (2). 

2.  What  is  the  equation  of  the  diameter  of  the  parabola 


which  bisects  the  chords  of  slope  ^  ? 

3.  Prove  Ex.  3,  §  1,  for  the  parabola. 

4.  There  are  no  conjugate  diameters  for  a  parabola. 


Why  ? 


6.  Extremities  and  Lengths  of  Conjugate  Diameters.  Ellipse. 
Let  the  coordinates  of  one  extremity  of  a  diameter  D  (not  an 
axis)  of  the  ellipse 


300 


ANALYTIC   GEOMETRY 


be  (»!,  yi).  .  The  slope  of  D  is  then  X  =  yi/ 
the  slope  X'  of  the  conjugate  diameter  D'  is 


From  (2),  §  2, 


«2A. 
Consequently,  the  equation  of  D'  can  be  written  in  the  form 

O  "a?"1""^2""1 

It  follows  that  D'  is  parallel  to  the  tangent  to  the  ellipse  at 

(x\>  Vi)-     In  other  words,  the  tan- 
v  gents  at  the  extremities  of  a  diam- 

eter are  parallel  to  the  conjugate 
\Xl>1/1     diameter. 

The  coordinates  of  the  ex- 
tremities of  D'  may  be  found  by 
solving  equations  (1)  and  (2) 
simultaneously.  -The  solutions 
are  found  to  be 


FIG.  9 


teA 

a) 


ayl        bxj\ 
b  '        a/ 


We  summarize  the  foregoing  results  in  a  theorem. 

THEOREM  7.  If  (xi,  yi)  is  one  extremity  of  a  diameter  D  of 
the  ellipse  (1),  then  (2)  is  the  equation  of  the  conjugate  diameter 
D',  and  one  extremity  (#/,  ?//)  of  D'  is 

b  '  1        a 

Suppose,  now,  that  we  denote  the  length  of  D  by  2  a±  and 
that  of  D'  by  2  b\.  Then  it  can  be  shown,  by  application  of 

(3)  and  the  equation  which  states  that  (xlt  3/1)  is  on  the  ellipse, 
that 

(4)  cd2  =  62  +  e2^2        and         b?  =  a2  —  e2^2. 
We  have,  then, 

ai2  +  &!2  =  a2  +  &2     or     (2  aj)2  +  (2  6i)2  =  (2  a)2  +  (2  6)2. 
This  result  we  express  as  a  theorem. 


DIAMETERS.     POLES  AND  POLARS 


301 


THEOREM  8.  TJie  sum  of  the  squares  of  the  lengths  of  any  two 
conjugate  diameters  of  an  ellipse  is  constant,  and  equals  the  sum 
of  the  squares  of  the  axes. 

Hyperbola.  Of  two  conjugate  diameters,  D  and  D',  of  the 
conjugate  hyperbolas 


(*)    («>  S-S-1- 


(6)  3-E--- 


one  meets  the  one  hyperbola ;  the  other,  the  other  hyperbola. 
Suppose  that  D  meets  (5  a)  and  D'  meets  (5  6),  and  that  the 
coordinates  of  an  extremity  of  D  on  (5  a)  are  (xlf  yj,  while 
those  of  an  extremity  of  D'  011  (5  6)  are  (a;/,  y/).  • 

Then  the  equations  of  D'  and  D  are,  respectively, 

as  is  evident,  from  analogy  to  the 
corresponding  equation  (2)  in  the 
case  of  the  ellipse.  From  (6)  it 
follows  that  the  tangents,  at  the 
extremities  of  a  diameter,  to  the 
hyperbola  on  which  these  extremi- 
ties lie  are  parallel  to  the  con- 
jugate diameter. 

The  coordinates  of  the  extremi- 
ties of  D'  can  be  found  by  solving  equations   (6  a)  and  (56) 
simultaneously.     The  solutions  are 


FlG  10 


^A         and         f_Sfc--.*a\ 

a  )  \       b  '        a  ) 


One  of  these  extremities  is  (a:/,  y/) ;  let  us  say,  the  first  one. 
Then  the  values  of  x/  and  y/  in  terms  of  xl  and  yl}  and  vice 
versa,  are 

6  a  6    '  a 

where  the  equations  (6)  are  obtained  by  solving  equations  (a) 
for  ON  and  w,. 


302  ANALYTIC   GEOMETRY 

The  student  should  note  the  symmetry  of  formulas  (6)  and 
(7)  in  (a5a,  yi)  and  (oV,  3^').  The  results  embodied  in  these 
formulas  we  state  as  a  theorem. 

THEOREM  9.  If  (xt,  y^)  is  one  extremity  of  a  diameter  D 
meeting  the  hyperbola  (5  a),  and  (#/,  y/)  is  a  properly  chosen  one 
of  the  extremities  of  the  conjugate  diameter  D',  then  (6)  are  the 
equations  of  D  and  D',  and  (7)  give  the  relations  between  the  two 
extremities. 

Let  2  ai  be  the  length  of  D  and  2  b±  that  of  D'.  It  can  be 
shown  that 

—  &2  =  eV2  +  a2, 


2        22  -a2  =  eV2  +  62. 
Hence  ax2  -  b{  =  a2  -  62 

and.  we  have  the  following  theorem. 

THEOREM  10.  The  difference  of  the  squares  of  the  lengths  of 
any  two  conjugate  diameters  of  a  hyperbola  is  constant,  and  equals 
the  difference  of  the  squares  of  the  axes. 

EXERCISES 

Establish  the  following  formulas. 

1.  Formulas  (3).  3.   Formulas  (7). 

2.  Formulas  (4).  4.   Formulas  (8). 

5.  Show  that  two  conjugate  diameters  of  a  hyperbola  are 
never  equal  unless  the  hyperbola  is  rectangular  and  that  in 
this  case  they  are  always  equal. 

6.  Prove  that  the  product  of  the  focal  radii  to  any  point  of 
an  ellipse  equals  the  square  of  half  the  diameter  conjugate  to 
the  diameter  through  the  point. 

7.  State   and    prove    the    corresponding    theorem    for    the 
hyperbola. 

7.  Physical  Meaning  of  Conjugate  Diameters.  Ellipse. 
Consider  a  flat  bar  of  iron,  on  the  end  ABCD  of  which  a 
circle  is  drawn.  Let  D  and  D'  be  any  two  mutually  perpen- 


DIAMETERS.     POLES  AND   POLARS  303 

dicular,    and    therefore    conjugate,    diameters    of    the    circle. 

Imagine  that  the  bar  is  subjected  to  heavy  pressure.     Then 

the  lengths  of  all  lines  parallel  to 

AB  and  CD  will  be  shortened  in  the 

same  ratio,  and  the  lengths  of  lines 

parallel  to  AD  and  BC  will  all  be 

lengthened  in  the  same  ratio ;   the  FIG.  n 

two    ratios    will    not,    however,    be 

equal.*     The  circle  will  thereby  be        jjiljjl 

carried  over  into  an  ellipse,  and  the 

diameters   D  and    D'   will    become 

conjugate  diameters  of  this  ellipse. 

A  proof  of  these  facts  will  be  given  Fio.  12 

shortly. 

The  student  can  perform  a  suggestive  experiment  by  taking 
an  ordinary  four-sided  eraser,  drawing  a  circle  and  the  diame- 
ters D,  D'  on  one  of  the  broader  faces  of  it,  and  then  pinching 
the  eraser  in  a  vise.  The  circle  will  go  over  into  an  oval  that 
looks  like  an  ellipse,  and  D  and  D'  will  remain  sensibly  straight 
lines. 

If  the  vise  is  set  too  hard,  the  bulging  will  be  considerable. 
But  imagine  the  ends  of  the  eraser  cut  off  square  and  the 
eraser  then  fitted  snugly  into  a  tube  or  chamber  of  rectangular 
cross-section,  with  the  broader  faces  and  the  ends  in  contact 
with  the -walls  of  the  chamber.  Let  the  chamber  be  closed  at 
one  end  by  a  rigid,  plane  diaphragm,  against  which  the  eraser 
is  to  be  pressed. 

If,  now,  a  plunger,  which  just  fits  the  chamber,  is  introduced 
and  pressed  down,  the  deformation  will  be  much  like  that  de- 
scribed in  the  opening  paragraph ;  the  circle  will  become  a 
true  ellipse,  and  D,  D',  remaining  straight  lines,  will  become 

*  Near  the  ends  AB  and  CD  of  the  cross-section  these  statements  will 
be  only  approximately  true,  since  there  will  be  a  slight  bulging  ;  and,  in- 
deed, there  will  also  be  a  slight  bulging  of  the  ends  themselves.  But  near 
the  middle  of  the  cross-section  the  deformation  will  be,  to  a  high  degree 
of  approximation,  as  described. 


304  ANALYTIC   GEOMETRY 

conjugate  diameters  of  the  ellipse.  But  there  will  be  one  es- 
sential difference,  in  that  in  the  first  case  lines  parallel  to  AD 
are  lengthened,  whereas  in  this  second  case  they  remain  un- 
changed ;  cf .  Figs.  11  and  12,  which  have  been  drawn  for  the 
second  case,  rather  than  for  the  first.  The  first  case  can,  how- 
ever, be  reduced  geometrically  to  the  second  if,  after  the  def- 
ormation has  been  made,  the  new  figure  is  reduced  in  scale, 
so  that  lines  parallel  to  AD  again  assume  their  original 
lengths. 

We  shall  confine  ourselves  to  the  second  case.  The  defor- 
mation of  the  plane  of  the  circle  may,  in  this  case,  be  called  a 
compression  in  one  direction  or  a  simple  compression.  All  line- 
segments  in  the  direction  of  compression  are  shortened  in  the 
same  ratio,  the  ratio  of  compression.  All  line-segments  in  the 
perpendicular  direction  remain  the  same  in  length ;  they  are 
all  moved  parallel  to  themselves,  with  the  exception  of  one 
which  remains  fixed.  In  the  case  described  this  one  rests 
against  the  diaphragm,  either  along  AD  or  BC.  If,  however, 
the  diaphragm  is  replaced  by  a  second  plunger,  the  fixed  line 
might  be  AD  or  BC  or  any  parallel  line  such  as  EF,  depending 
on  the  manner  in  which  the  pressures  on  the  two  plungers  are 
applied.  This  line,  perpendicular  to  the  direction  of  compres- 
sion and  having  all  its  points  fixed  under  the  compression,  we 
shall  call  the  central  line. 

In  studying  the  effects  of  a  compression  let  us  take  the  cen- 
tral line  as  the  axis  of  x  and  the  ratio  of  compression  as  Z; 
I  is  a  positive  constant  <  1.  We  prove  first  that  the  compres- 
sion carries  a  straight  line  L  into  a 
straight  line  L.  This  is  obvious  if  L 
is  parallel  to  either  axis.  If  L  is  any 
other  line,  the  similar  triangles  in 
Fig.  13  show  that  it  goes  over  into  a 
F  line  L,  and  that  if  L  is  of  slope  \,  L  is 

of  slope  IX. 

Next,  consider  an  arbitrary  circle,  with  center  0  (Fig.  1'4). 
The  diameter  A  A  of  the  circle  which  is  parallel  to  the  central 


DIAMETERS.     POLES  AND   POLARS 


305 


V,Y,y 


line,  the  axis  of  x,  goes  over  into  a  parallel  line-segment,  A' A, 
of  equal  length,  and  the  mid-point,  0,  of  A'A  goes  over 
into  the  mid-point,  0',  of  A'A. 

Let  the  circle  be  referred  to 
axes  (oj,  y)  with  the  origin  at  O, 
the  axis  of  x  lying  along  A'A. 
Its  equation  will  be 

(1) 


3?  +  y*  =  a2. 

Let  the  curve  into  which  the  circle 
is  deformed  be  referred  to  axes 
(X,  Y)  with  the  origin  at  0',  the 
axis  of  X  lying  along  A'A.  Then 
any  point  (x,  y)  on  the  circle  goes  over  into  a  point  (X,  Y) 

such  that 

Y 
X=x,         Y=ly,  or  x=X,         */=-• 


I 


It  follows,  then,  that  the  circle  (1)  is  transformed  into  the 
curve 


or 

(2)  1 =1,  b  =  la. 

a?       &2 

Thus  the  circle  is  seen  to  be  carried  into  an  ellipse. 

It  remains  to  prove  that  the  lines  D  and  D',  into  which  two 
conjugate  diameters  D  and  D'  of  the  circle  are  carried  by  the 
compression,  are  conjugate  diameters  of  the  ellipse.  If  the 
angle  <£  is  as  shown  in  Fig.  14,  the  slopes  of  D  and  D'  are 
tan  <£  and  tan(<£  +  90°)=  —  cot<£.  Hence  the  slopes  of  D 

and  D'  are 

X  =  I  tan  <£        and        X'  =  —  I  cot  <f>. 

Then  AX'  =  - 12  tan  0  cot  <£  =  -  I2, 

or,  since,  by  (2),  I  =  b/a, 

XX'  =  -  -. 


306 


ANALYTIC   GEOMETRY 


p't 


FIG.  15 


Consequently,  D  and  D',  according  to  Th.  3,  §  2,  are  conjugate 
diameters  of  the  ellipse,  q.  e.  d. 

If  the  center  0  of  the  circle  lies  on  the  central  line  (Fig.  15), 
the  circle  is  the  auxiliary  circle  of  the  ellipse  (Ch.  VII,  10), 

and  the  angle  <f>  is  the  eccentric 
angle  for  the  extremity  P  of  the 
diameter  D.  The  eccentric  angle 
for  the  extremity  P'  of  the  con- 
jugate diameter  D'  is,  clearly, 
<£  +  90°,  or  <£  +  n90°,  where  n  is 
an  odd  number.  Consequently,  we 
have  proved  the  following  theorem. 
THEOKEM  11.  The  eccentric  angles 
for  two  points  of  an  ellipse  which  are 
extremities  of  two  conjugate  diame- 
ters differ  by  90°,  or  by  an  odd  multiple  of  90°. 

The  theorem  is  essentially  the  geometrical  equivalent  of  the 
physical  property  of  conjugate  diameters  which  we  have  been 
discussing.  It  furnishes  a  method  of  constructing  rapidly  as 
many  pairs  of  conjugate  diameters  of  an  ellipse  as  may  be 
desired. 

The  parametric  representation  of  the  ellipse  can  be  used  to 
great  advantage  throughout  the  study  of  conjugate  diameters. 
The  extremity  P  of  the  diameter  D  (Fig.  15)  has,  by  Ch.  VII, 
§  10,  the  coordinates 

(3)  x  =  a  cos  <£,  y  =  b  sin  <£. 

Then  the  extremity  P'  of  the  conjugate  diameter  D'  has,  by 
Th.  11,  the  coordinates 

(4)  xr  =  —  a  sin  </>,  y'  =  b  cos  <f>. 

Hence  we  obtain,  for  the  squares  of  the  half-lengths  of  D  and 
D': 

aj2  =  d?  cos2  <f>  +  62  sin2  <f>,  b?  =  a2  sin2  0  +  i2  cos2  <£. 

Therefore,  af  +  b?  =  a2  +  62, 

and  we  have  a  simple  proof  of  Th.  8,  §  6. 


DIAMETERS.     POLES  AND   POLARS 


307 


Hyperbola.  Consider,  now,  a  set  of  four  steel  girders  A,  B, 
C,  and  D  in  the  form  of  a  square,  and  a  cross  girder  E  through 
the  center  of  the  square  parallel  to  A  and  C.  Suppose  that 
tie-rods  are  spanned  into  the  frame  along  the  diagonals  of 
the  square  and  along  pairs  of  lines  making  equal  angles  with 
the  diagonals.  These  pairs  of  tie-rods, 
then,  lie  along  conjugate  diameters  of  a 
rectangular  hyperbola,  of  which  the 
diagonals  of  the  square  are  the  asymp- 
totes (Th.  6,  §  4). 

Suppose  that  the  girder  E  is  firmly 
set  in  masonry,  so  that  it  is  immovable, 
and  suppose  that  equal  tensions  are 
exerted  on  the  girders  A  and  C  as 

shown.  Then  the  square  is  elongated  into  a  rectangle,  except 
for  a  slight  bulging ;  the  diagonal  rods  come  to  lie  along  the 
diagonals  of  the  rectangle  and  the  other  pairs  of  tie-rods  take 
on  the  positions  of  pairs  of  lines  which  are  conjugate  di- 
ameters in  a  hyperbola  having  the  diagonals  of  the  rectangle 
as  asymptotes,  as  we  shall  presently 
show. 

In  this  case  we  speak  of  an  elongation 
in  one  direction  or  a  simple  elongation. 
The  line  of  the  girder  E  is  the  central 
line  of  the  elongation,  and  the  ratio 
I  ( >  1),  in  which  all  distances  perpen- 
dicular to  E  are  stretched,  is  the  ratio  of 
elongation. 

Let  us  take  the  central  line  as  axis  of  x. 
Since  the  slope  of  the  diagonal  S  of  the 
square  is  1,  the  slope  of  the  diagonal  S  of  the  rectangle  is  I. 
If  thejmgle  fa  is  as  shown,  Fig.  16,  the  slopes  of  the  two 
lines  D  and  D'  making  equal  angles  with  S  are  tan  fa  and 
tan  (90  —  fa)—  cot  fa.  JHence  the  slopes,  A  and  A.',  of  the  lines 
D  and  D',  into  which  D  and  D'  are  carried,  are 
A  =  I  tan  fa        and         A'  =  I  cot  fa. 


t  t 


308  ANALYTIC   GEOMETRY 

But  then  XX'  =Z2, 

and,  according  to  Th.  5,  §  4,  the  lines  D  and  D'  are  conjugate 
diameters  of  a  hyperbola,  of  which  the  diagonals  of  the  rec- 
tangle are  the  asymptotes.  The  ratio  b/a  of  the  axes  of  the 
hyperbola  equals  the  ratio  of  elongation,  I  : 


Finally,  let  us  show  that  the  elongation  carries  the  rectangu- 
lar hyperbola 
/  K\ y  ^^  "i 

W  tf~tf 

Ct  Ur 

into  the  hyperbola 

(6)  *?_#!=  i. 

a2     62 

The  two  hyperbolas  have  the  same 
auxiliary  circle,  and  the  same  eccentric 
angle,  fa,  for  points,  P :  (a^,  y^)  and 
P :  (a?i,  7/i),  with  the  same  abscissa. 
Hence,  according  to  the  method  of 

parametric  representation  of  a  hyperbola  (Ch.  VIII,  §  9),  the 

coordinates  of  P  and  P  are 


FIG  is 


(5  a) 
(6  a) 

Therefore 


=  a  sec  fa, 
—  a  sec  fa, 


yl  =  a  tan  fa ; 
yi  =  b  tan  fa. 


or,  since  b/a  =  1,  yi  =  fiji. 

Hence  the  elongation  does  carry  the  hyperbola  (5)  into  the  hy- 
perbola (6). 

Let  P,  with  coordinates  (6  a),  be  an  extremity  of  the  diam- 
eter D.  Then  the  coordinates  of  an  extremity  P'  of  the  conju- 
gate diameter  D'  are,  by  (7),  §  6, 


(7) 


i  =  —  =  b  sec  fa. 


DIAMETERS.     POLES  AND   POLARS  309 

Here  again,  then,  the  use  of  the  eccentric  angle  gives  sym- 
metry to  the  results. 

One-Dimensional  Strains.  In  the  case  of  the  ellipse  we 
might  equally  well  have  subjected  the  given  circle  to  an  elon- 
gation, and  in  the  case  of  the  hyperbola  we  might  have  com- 
pressed the  given  equilateral  hyperbola,  instead  of  elongating  it. 
Compression  and  elongation  in  one  direction  are  but  two  types 
of  a  single  kind  of  deformation,  known  as  a  one-dimensional 
strain.  If  the  coefficient  I  of  the  strain  is  greater  than  unity, 
the  strain  is  an  elongation ;  on  the  other  hand,  if  I  <  1,  the 
strain  is  a  compression. 

EXERCISES 

1.  Repeat  Ex.  1  of   §  2,  drawing  the  auxiliary  circle  and 
constructing  the  diameters  conjugate  to  the  given  diameters 
by  application  of  Theorem  11. 

2.  Draw  in  pencil  the  asymptotes  and  a  number  of  pairs  of 
conjugate   diameters,   including   the   axes,   of    a   rectangular 
hyperbola.     Construct  in  ink  the  lines  into  which  the  given 
lines  are  carried  by  the  compression  of  ratio  -f   which   has 
an  axis  of  the  hyperbola  as  central  line.     What  does  the  re- 
sulting figure  represent  ? 

3.  Prove  Th.  10,  §  6,  by  means  of  formulas  (6  a)  and  (7)  of 
the  present  paragraph. 

8.  Harmonic  Division.  Let  P^  be  a  line-segment,  and  let 
Qx  be  one  of  its  points.  Then  Q±  divides  PiP2  internally  in  a 
certain  ratio,  p.  (Ch.  I,  §  6) : 

PiQi=  ~~% PI  4'     3~~ 

QiP2      **'  FIG.  19 

On  PiPz  produced  construct  the  point  Q2  which  divides 
externally  in  the  same  ratio : 


310  ANALYTIC   GEOMETRY 

The  points  Qi  and  Q.z  are  said  to  divide  the  segment  PXP2 
harmonically;  they  divide  PiP2  internally  and  externally  in 
the  same  ratio : 

a)  ifhtpr 

Let  us  start  with  /*  as  given  and  trace  the  changes  in  Qx  and 
Q2  as  fji  varies.  If  p.  =  0,  Qi  and  Q2  coincide  in  P^  As  p.  in- 
creases from  0  to  1,  Qi  moves  to  the  right  from  P:  to  the  mid- 
point M  of  PiP2,  and  Q2  moves  to  the  left  and  recedes  indefi- 
nitely. If  fj.  =  1,  Qi  is  at  M ;  but  Q2  has  disappeared.  Thus 
there  is  no  point  which,  with  the  mid-point  of  a  segment,  divides  the 
segment  harmonically.  As  /u,  increases  from  1  without  limit,  Qt 
proceeds  from  M  toward  P2  as  its  limit,  and  Q2  appears  again 
from  the  extreme  right,  continually  moving  in  and  approach- 
ing P2  as  its  limit. 

The  proportion  (1)  may  be  written  in  the  form : 


this  new  proportion  says  that  Pl  and  P2  divide  the  segment 
QiQz  harmonically.     Thus  we  have  the  following  theorem. 

THEOREM  1.  If  the  points  Qi  and  Q2  divide  the  line-segment 
P^Pz  harmonically,  then,  reciprocally,  the  points  Px  and  P2  divide 
the  line-segment  QiQ2  harmonically. 

In  other  words,  the  relationship  between  the  two  pairs  of 
points  is  symmetric. 

Suppose  that  P!  and  P2  have  the  coordinates  (a^,  y^  and 
(cc-j,  y2).  Then  the  coordinates  (xi,  t//)  and  (a^',  y2')  of  Qx  and 
Q2  are  given  by  formulas  (1)  and  (2)  of  §  6,  Ch.  I.  If,  in 
each  of  these  formulas,  we  divide  the  numerator  and  denomi- 
nator by  ra2  and  then  set  mi/m?,  =  p.,  we  obtain,  as  the  desired 
coordinates : 


1+f. 


<*'.: 


DIAMETERS.     POLES  AND   POLARS 


311 


EXERCISES 

1.  Four  points  Px,  P2,  Q1}  Q2  on  the  axis  of  x  have,  respec- 
tively, the  abscissas  3,  8,  5,  —  7.     Show  that  Qlt  Q2  divide  PjP2 
harmonically,  and  find  the  common  ratio  p.  of  internal  and  ex- 
ternal division.     Find,  also,  the  value  of  the  ratio,  //,  for  the 
division  by  PI  and  P2  of  the  segment  QiQ2. 

2.  Find  the  point  on  the  axis  of  x  which,  with  the  point 
(—1,  0),  divides  harmonically  the  segment  of  the  axis  joining 
the  points  (-8,0),  (3,  0). 

3.  Exercise  1,  for  the  four  points  Pj,  P2,  Qlf  Q2  with  the 
respective  coordinates  (2,  3),  (—  1,  9),  (1,  5),  (5,  —3). 

4.  Find  the  point  which,  with  the  point  (2,  1),  divides  har- 
monically the  line-segment  joining  the  points  (5,  —  2),  (1,  2). 

9.   Polar  of  a  Point.     Consider  the  following  locus  problem. 
The  ellipse 

*  L 


P:(X.Y) 


FIG.  20 


and    the    point    Pl :  (xl}  yt)    are 

given.    A  line  L  is  drawn  through 

PX  meeting  the  ellipse  in  Qt  and 

Q2,  and  on  L  the  point  P :  (X,  Y)    • 

is  marked  which,  with  P1?  divides 

QiQ2  harmonically.      What  is  the  locus  of  P,  as  L  revolves 

about PX? 

Since   Plf  P  divide  QiQ2  harmonically,  Qif  Q2  divide  P^P 
harmonically.     Hence  the  coordinates  (#/,  y/),  (x2,  y2)  of  Ql} 


are  : 


tSfm 


T3   ~~' 

1+/X 


. 
i- 


As  L  rotates,  the  ratio  p.  varies ;  it  is,  then,  an  auxiliary 
variable  expressing  analytically  the"  rotation  of  L. 


312  ANALYTIC   GEOMETRY 

The  coordinates  of  Q1  and  Q2  satisfy  (1).  Substituting  them 
in  turn  in  (1)  and  clearing  each  of  the  resulting  equations  of 
fractions,  we  have 


To  eliminate  /A,  we  subtract  the  second  equation  from  the  first, 
thus  getting 

F  =  4  /.xa262, 


or,  finally,  +        =  1. 

a2        o2 

The  locus  of  P  is,  therefore,  a  straight  line,  or  a  portion  of 
a  straight  line.*  This  line  is  known  as  the"  polar  of  the  point 
P!  with  respect  to  the  ellipse.  Hence  we  may  say  : 

TJie  polar  of  the  point  (xi}  y^  with  respect  to  the  ellipse  (1)  has 
the  equation 

(2}  ^  +  M=i 

a2       62 

This  equation  is  identical  in  form  with  the  equation  /)f  the 
tangent  to  the  ellipse  at  the  point  (xl}  y^),  Ch.  IX,  §  2,  (12). 
But  in  the  present  problem  (a^,  y^)  is,  in  general,  not  on  the 
curve,  and  then  (2)  represents  a  line  which  is  not  a  tangent. 

If,  in  particular,  Pl  :  (a^,  y±)  is  on  the  ellipse,  then  (2)  does 
represent  the  tangent  at  P!.  Accordingly,  we  should  like  to 
say  :  The  polar  of  a  point  on  the  ellipse  is  the  tangent  at  the 
point.  Now  there  is  trouble,  geometrically,  when  P±  is  on  the 
ellipse.  For  then  Ql  or  Q»  coincides  with  PI,  and  P  coincides 
with  them,  so  that,  actually,  no  polar  is  defined.  Suppose, 
however,  that  P!  is  a  point  near  to  P1}  but  not  on  the  curve. 
Then  it  can  be  shown  (  Exs.  1,  2)  that  the  limiting  position 
of  the  polar  of  Pls  when  P:  approaches  Px  as  its  limit,  is  the 
tangent  at  Px.  Hence  the  above  statement  is  substantiated, 

*  If  PI  is  inside  the  coni<5,  the  locus  is  the  entire  line,  but  if  PI  is  out- 
side the  conic,  the  locus  consists  of  only  those  points  of  the  line  which 
are  inside  the  conic. 


DIAMETERS.     POLES  AND   POLARS  313 

not  as  a  conclusion,  but  as  a  proper  definition  of  the  polar  of 
a  point  on  the  ellipse. 

If  P!  is  at  the  origin,  it  is  always  the  mid-point  of  QiQ2,  and 
so  there  is  never  a  point  P  which,  with  Pl}  divides  Q^Q^  har- 
monically. Consequently,  the  origin  has  no  polar. 

The  foregoing  discussion  is  valid  for  the  hyperbola 


if  in  the  equations  we  replace  6-  by  —  &2.  Thus,  the  polar  of 
the  point  (x1}  y^)  with  respect  to  the  hyperbola  (3)  has  the  equation 

(A\  £!£_M.=  1 

a?       62 

The  polar  of  a  point  on  the  hyperbola  is  defined  as  the  tangent 
at  the  point.  The  center  of  the  hyperbola  has  no  polar. 

We  can  now  state  the  following  theorem. 

THEOREM  2.  Given  a  central  conic,  C.  Every  point  in  the 
plane,  except  the  center  of  C,  has  a  polar  with  respect  to  C. 

Let  the  student  show  that  the  polar  of  the  point  (xi}  y^)  with 
respect  to  the  parabola 

(5)  y*  =  2mx 
has  the  equation 

(6)  yjy=m(x+xl'). 

If  we  define  the  polar  of  a  point  on  the  parabola  as  the  tangent 
at  the  point,  equation  (6)  shows  that  there  are  no  exceptions 
in  this  case.  Accordingly,  we  have  the  theorem  : 

THEOREM  3.  Every  point  in  the  plane  has  a  polar  with  re- 
spect to  a  parabola. 

From  the  definition  of  a  polar  it  is  evident  that  the  polar  of 
a  point  internal  to  a  conic  does  not  cut  the  conic,  and  that  the 
polar  of  a  point  external  to  a  conic  does  cut  the  conic.  In  the 
intermediate  case,  when  the  point  lies  on  the  conic,  the  polar 
is  a  tangent. 


314 


ANALYTIC   GEOMETRY 


FIG.  21 


EXERCISES 

1.    Show  that  the  polar  of  a  point  PI  external  to  a  conic  is 
the  line  LI  drawn  through  the  points  of  contact  of  the  tangents 
to  the  conic  from  P^ 

.Suggestion.  Prove  that  P  (Fig.  21) 
approaches  K  as  its  limit,  when  the  line 
L,  rotating  about  P1}  approaches  T. 

2.  Prove  that,  if  the  point  PI  of  Ex.  1 
approaches  a  point  on  the  conic  as  its 
limit,  then  its  polar,  LI,  will  approach 
the  tangent  at  this  point. 

3.  Establish  formula  (6). 

In  each  of  the  following  exercises,  find  the  equation  of  the 
polar  of  the  given  point  with  respect  to  the  given  conic  and 
draw  a  figure,  showing  the  conic,  point,  and  polar. 

Conic  Point 

4.  x*  +  tf  =  9,  (0,2). 

5.  3x-  +  5y2  =  15,  (5,6). 

6.  a5*-2f=16,  (2,1). 
•7.   2y*-5x  =  Q,  (-3,4). 

8.  Prove  that  in  any  conic  the  polar  of  a  focus  is  the  cor- 
responding directrix. 

9.  Prove  that  the  polar  of  a  point  Px  with  respect  to  a 
circle,  center  at  0,  is  perpendicular  to  the  line  OP^ 

10.  Show,  further,  that  the  product  of  the  distances  of  0 
from  P1  and  the  polar  of  Pt  is  the  square  of  the  radius  of  the 
circle. 

11.  On  the  basis  of  the  results  of  Exs.  9,  10,  discuss  the 
variation  in  position  of  the  polar  of  a  point  P  with  respect  to 
a  circle,  (a)  when  P  moves  on  a  straight  line  through  the 
center  of  the  circle;    (6)  when  P  traces  a  circle,  concentric 
with  the  given  circle. 


DIAMETERS.     POLES  AND   POLARS  315 

Find  the  equation  of  the  polar  of  the  point  (x1}  y^)  with  re- 
spect to  each  of  the  following  conies. 

12.  The  hyperbola  :     xy  =  k. 

13.  The  circle  :     (a  —  a)2  +  (y  —  /3)2  =  pz. 

14.  The  conic  :     (1  —  e2)  a2  +  y2  —  2  mx  +  m2  =  0. 

10.  Pole  of  a  Line.  If,  with  respect  to  a  given  conic,  the 
line  L  is  the  polar  of  the  point  P,  the  point  P  is  known  as  the 
pole  of  the  line  L. 

Given  a  conic  and  a  line  L  ;  to  find  the  pole,  P,  of  L  with 
respect  to  the  conic. 

Let  the  conic  be  the  ellipse, 

(1)  2*2+82,2  =  6, 
and  L,  the  line, 

(2)  4z-3y-2  =  0. 

If  we  denote  the  coordinates  of  P  by  (xif  3^),  the  polar  of  P 
with  respect  to  (1)  is 

(3)  20^+3^-6=0. 

But  the  polar  of  P  was  given  as  the  line  (2).  Equations  (2) 
and  (3),  then,  represent  the  same  line.  Consequently,  by 
Ch.  II,  §  10,  Th.  5, 


4        -3      -2 
Then  xl  =  6,  yt  =  -  3 

and  so  the  point  (6,  —  3)  is  the  pole  of  the  line  (2)  with  re- 
spect to  the  ellipse  (1). 

We  now  raise  the  question:  Has  every  line  a  pole  with 
respect  to  a  given  conic  ?  Let  us  answer  this  question  first 
for  the  central  'conies.  Equations  (2)  and  (4)  of  §  9,  which 
represent  the  polars  of  a  given  point  with  respect  to  the  central 
conies  (1)  and  (3)  of  §  9,  are  never  satisfied  by  x  =  0,  y  =  0, 
no  matter  where  the  given  point  lies.  Consequently,  the  polar 


316  ANALYTIC   GEOMETRY 

of  a  point  with  respect  to  a  central  conic  never  passes  through 
the  center  of  the  conic.  In  other  words,  a  diameter  of  a  central 
conic  has  no  pole. 

We  proceed  to  show  that  every  other  line  has  a  pole,  giving 
the  proof  in  the  case  of  the  hyperbola 

(4)  £-£=1. 

a2      Z>2 

Any  line  not  a  diameter  of  (4),  that  is,  not  passing  through 
the  origin,  can  be  represented  by  an  equation  of  the  form  * 

(5)  Ax  +  By  =  1. 

If  (a?!,  2/1)  is  the  pole  of  this  line,  the  line  also  has  the  equation 

xix    y\y  _  i 

a2        62 

Hence  ^:  A  =  -^:B  =  1  : 1, 

a2  62 

and  xv  =  a2  A,  y\  =  —  o2B. 

We  see,  then,  that  the  line  (5)  has  always  a  definite  pole, 
namely,  the  point  (a2 A,  —  bzB),  q.  e.  d. 

In  the  case  of  the  ellipse  the  proof  is  similar. 

As  regards  the  parabola, 

y2  =  2  mx, 

the  equation  of  the  polar  of  (a^,  y^) : 

y$  =  mx  +  ma?!, 

has  one  term  which  can  never  drop  out,  no  matter  where  fa,  y^) 
lies,  — namely,  the  term  mx.  Thus  the  polar  can  never  be 
parallel  to  the  axis  of  x,  or  coincide  with  it.  In  other  words, 
a  diameter  of  a  parabola  has  no  pole.  It  can  be  shown,  how- 
ever, that  every  other  line  has  a  pole.  The  proof  is  left  to  the 
student ;  cf.  Ex.  1. 

The  foregoing  results  we  now  summarize  in  the  form  of  a 
theorem. 

*  A  and  B  are  not  both  zero. 


DIAMETERS.     POLES  AND   POLARS  317 

THEOREM  4.  Given  a  conic  C.  Every  line  of  the  plane, 
which  is  not  a  diameter  of  C,  has  a  pole  with  respect  to  C. 

By  comparing  this  theorem  with  Theorems  2,  3,  §  9,  we 
see  that  the  lines  which  have  no  poles  with  respect  to  a  conic 
go  through  the  point  which  has  no  polar,  provided  these  lines 

intersect. 

EXERCISES 

1.  Give  the  proof  of  Theorem  4  for  the  parabola. 

In  each  of  the  following  exercises  find  the  pole  of  the  given 
line  with  respect  to  the  given  conic. 

Conic  Line 

2.  a?  +  ty«  =  8,  2x-3y-2  =  Q. 

3.  5 a2  -  6?/2  -30  =  0,  4a  +  2y-7  =  0. 

4.  3yn~-Sx  =  Q,  2 a- 3  =  0. 

5.  Itf-  +  2y*=U,  6x  +  5y-8  =  Q. 

6.  Prove  that  the  pole  of  any  line  through  the  focus  of  a 
conic  is  a  point  on  the  corresponding  directrix. 

7.  Given  the  circle  x-  +  y1  =  a2.     Prove  that  the  pole^  with 
respect  to  this  circle,  of  a  line  moving  so  that  it  is  always 
tangent  to  a  concentric  circle  traces  a  second  concentric  circle. 
Cf.  Exs.  9-11,  §  9. 

11.  Properties  of  Poles  and  Polars.  The  poles  and  polars  * 
discussed  in  this  paragraph  are  all  taken  with  reference  to  an 
arbitrarily  given  conic.  For  the 
sake  of  brevity  mention  of  the 
conic  is,  in  general,  suppressed. 

THEOREM  5a.     If  a  point  Pv  lies 
on  the  polar  of  a  second  point  P2, 

then,  conversely,  P2  lies  on  the  polar 

FIG.  22 
of  PI. 

Let  the  polats  of  the  points  Pt  and  P2  be  LI  and  L2.  Then 
the  theorem  says  that,  if  P1  lies  on  Z/2,  P2  lies  on  L^  But  this 

*  Only  those  points  which  have  polars  and  those  lines  which  have  poles 
are  considered. 


318  ANALYTIC   GEOMETRY 

is  the  same  as  saying  that,  if  £2  goes  through  P1}  L±  goes 
through  P2,  °r  vice  versa.  This  second  form  of  the  statement 
we  enunciate  as  a  theorem. 

THEOREM  56.  If  a  line  Li  goes  through  the  pole  of  a  second 
line  L2,  then,  conversely,  L2  goes  through  the  pole  of  Lt. 

Since  the  two  theorems  are  equivalent  in  content,  and  differ 
only  in  point  of  view,  a  proof  of  one  also  proves  the  other. 
We  choose  to  prove  Theorem  5a,  and  to  give  the  proof  in  the 
case  that  the  given  conic  is  the  hyperbola 


a2     62 

The  proofs  in  the  other  two  cases  are  similar. 

Let  PI  and  P2  have  the  coordinates  (x1}  y^)  and  (a^,  ?/2). 
Then  Ll  and  L2  have  the  equations 

Ei?_M  =  l        and         ^^M=l 
a2       62  ~  a2       ft2 

The  cpndition  that  Pt  lies  on  L2  is 


a2        62  " 
and  the  condition  that  P2  lies  on  LI  is 


a2 


But  these  two  conditions  are  the  same.  Hence,  if  Pj  lies  on 
L2,  then  P2  lies  on  LI,  and  conversely,  q.  e.  d. 

Suppose,  now,  that  we  join  the  points  P!  and  P2  of  "Fig.  22 
by  the  line  L.  Since  P!  lies  on  L,  it  follows,  by  Th.  5cr,  that 
the  pole,  P,  of  L  lies  on  Lv.  Similarly,  since  P2  lies  on  L,  P 
lies  also  on  L».  Hence,  P  is  the  point  of  intersection  of  Lt  and 
L2.  Thus  we  have  the  theorem  : 

THEOREM  6a.  The  pole  of  the  line  joining  two  points,  P^  and 
P2,  is  the  point  of  intersection  of  the  polars  of  Pl  and  P2. 

Starting  again,  we  bring  the  lines  LI  and  L2  of  Fig.  22  to 
intersection  in  P.  By  Th.  56,  since  each  of  the  lines  Lt  and 


DIAMETERS.     POLES  AND   POLARS 


319 


FIG.  23 


LZ  goes  through  P,  it  follows  that  the  polar,  L,  of  P  goes 
through  each  of  the  poles,  PI  and  P2,  of  LI  and  L».  Conse- 
quently, L  is  the  line  joining  PI  and  P2  and  we  have  proved 
Theorem  66 : 

THEOREM  66.  Tlie  polar  of  the  point  of  intersection  of  two 
lines,  LI  and  Lz,  is  the  line  joining  the  poles  of  LI  and  L& 

By  application  of  either  Ths.  5a,  56  or  Ths.  6a,  66,  the 
student  can  easily  prove  the  following  theorems. 

THEOREM  la.  If  a  number  of  points 
all  lie  on  a  line,  L,  their  polars  all  go 
through  a  point,  namely,  the  pole  of  L. 

THEOREM  76.     If  a  number  of  lines  all 

go  through  a  point,  P,  their  poles  all  lie  on 

a  line,  namely,  the  polar  of  P. 

Finally,  take  a  line  L  which  cuts  the 

given  conic  in  two  points,  P^  and  P2. 

Since  L  is  the  line  joining  Pl  and  P2)  the  pole,  P,  of  L  is  the 
point  of  intersection  of  the  polars  of  PI 
and  P2  (Th.  6a),  that  is,  of  the  tangents 
to  the  conic  at  Pt  and  P2-  Thus  we  have 
proved  the  theorem : 

THEOREM  8a.     The,  pole,  of  a  line  inter- 
secting the   given   conic  in   two  points,  Pl 
and  P2,  is  the  point  of  intersection  of  the 
tangents  to  the  conic  at  P±  and  P.2. 

Let  the  student  prove  the  mate  of  this  theorem,  namely : 
THEOREM  86.     The  polar  of  a  point  external  to   the  given 
conic  is  the  line  joining  the  points  of  contact  of  the  tangents  to  the 
conic  from  the  point. 

Theorem  8a  furnishes  a  means  of  constructing  the  pole  of 
a  line  which  meets  the  given  conic  ;•  Theorem  86,  a  means  of 
constructing  the  polar  of  a  point  external  to  the  conic. 

To  construct  the  pole,  P,  of  a  given  line,  L,  which  does  not 
meet  the  conic  (Fig.  25),  choose  any  two  points,  Px  and  P2,  on 


FIG.  24 


320 


ANALYTIC   GEOMETRY 


FIG.  25 


L,  and  construct  their  polars,  Z/j  and  Lz. 
Since  L  is  the  line  joining  Pt  and  P2,  its 
pole,  P,  is  the  point  of  intersection  of  L± 
and  Z/2. 

The  student  should  now  establish  the 
analogous  construction  for  the  polar  of 
a  given  point  which  is  internal  to  the 
conic.* 


EXERCISES 

1.  Prove  Theorems  7a,  76. 

2.  Prove  Theorem  86. 

3.  Show  how  to  construct  the  polar  of  a  given  point  which 
is  internal  to  the  conic.     Prove  the  validity  of  the  construction. 

4.  On  the  basis  of  Theorem  86  develop  in  detail  a  method 
for  finding  the  equations  of  the  tangents  to  a  conic  from  an 
external  point. 

By  means  of  this  method  find  the  equations  of  the  tangents 
required  in  each  of  the  following  exercises  of  Ch.  IX,  §  7. 

5.  Exercise  5.  6.   Exercise  6. 
7.   Exercise  9.                                8.   Exercise  12. 

12.  Relative  Positions  of  Pole  and  Polar.     Central  Conies. 
The    following   theorem   is   in- 
structive  concerning    the    rela- 
tive positions  of  pole  and  polar 
with  regard  to  a  central  conic. 

THEOREM  9.  Let  the  point  P± 
and  the  line  LI  be  pole  and  polar 
in  a  central  conic,  center  at  0  ; 
let  D  be  the  diameter  through  FIG.  26 

*  These  methods  are  not  very  serviceable  if  accurate  constructions  are 
desired,  since  they  involve  the  construction,  not  only  of  the  tangent  at  a 
given  point  of  the  conic,  but  also  of  the  tangents  from  an  external  point ; 
cf .  §  13.  They  are,  however,  useful  in  rough  work. 


DIAMETERS.     POLES  AND  POLARS 


321 


PI,  D'  the  diameter  conjugate  to  D,  and  Pz  the  point  of  inter- 
section of  Li  with  D.  Then  LI  is  parallel  to  D'  and  the  half- 
length,  d,  of  D  is  a  mean  proportional  between  OP\  and  OP2  : 

(1)  OP1.OP2  =  d2. 

We  will  prove  the  second  part  of  the  theorem  first.  By  the 
definition  of  the  polar  of  a  point,  PjP2  is  divided  harmonically 
by  the  points,  Qi  and  Qz,  in  which  D  meets  the  conic.  Conse- 
quently, by  (1),  §  8, 


Expressing  each  of  the  four  distances  in  terms  of  OP\,  0P2, 
and  OQi  =  OQ2  =  d,  we  have 

(OP1  -  d)(OP2  +  d)  =  (0Pi  +  d)(d  -  OP2). 

On   multiplying  out  and   reducing,  we   obtain   equation   (1), 
q.  e.  d. 

We  will  give  the  proof  of  the  first  part  of  the  theorem  in 
the  case  that  the  conic  is  the  ellipse 


(f     F 

Let  the  coordinates  of  Pj  be  fa,  y^)  and  those  of  Q1}  fa,  y2). 
Then  the  equations  of  LI  and  D'  are,  respectively,  by  §  9;  (2), 
and  §  6,  (2), 

/O\  •k\X     .     y \jj          -t  T  *t/o*C     .     ylV          /\ 

( 6 )  h  ^^^  =  ±         ana        — = — (-  ^^^  =  u. 

a2       62  a2       62 

Since  Px  and  Qx  are  on  a  line  with  0,  their  coordinates  are 
proportional : 

and  hence  so  are  the  left-hand  sides  of 
equations  (2).  Consequently,  L±  and^' 
are  parallel,  q.  e.  d. 

The  proof  of  the  second  part  of  the 
theorem  assumes  that  D  meets  the  conic. 
This  is  not  true,  however,  if  the  conic  is 
a  hyperbola  and  Px  lies  in  an  opening  FIG.  27 


v   /D 


322 


ANALYTIC  GEOMETRY 


between  the  asymptotes  not  containing  a  branch  of  the  hyper- 
bola. In  this  case,  too,  the  theorem  is  valid,  but  the  points 
PI  and  P2,  instead  of  being  on  the  same  side  of  0,  are  on 
opposite  sides.* 

COROLLARY.  The  points  Px  and  P2  are  on  the  same  or  oppo- 
site sides  of  0,  according  as  D  meets  or  does  not  meet  the  conic. 

Given,  now,  a  point,  P,  and  its  polar,  L,  with  respect  to  a 
central  conic.  If  P  traces  a  diameter  D,  then  L  moves  always 
parallel  to  the  conjugate  diameter. 

In  the  case  that  D  intersects  the  conic,  and  P  is  an  intersec- 
tion, L  is  the  tangent  at  P.  If  P  then  moves  in  along  D 
toward  the  center  as  its  limit,  L  ceases  to  meet  the  conic,  and 
recedes  indefinitely.  On  the  other  hand,  if  P  moves  out 
along  Z),  receding  indefinitely,  L  moves  in  toward  the  center, 
and  approaches  the  diameter  conjugate  to  D  as  its  limit. 

The  case  in  which  the  conic  is  a  hyperbola  and  D  intersects 
the  conjugate  hyperbola  remains.  If  P  is  at  one  of  the  inter- 
sections, L  is  the  tangent  to  the  conjugate  hyperbola  at  the 
other ;  cf .  Ex.  6.  If  P  then  moves  in  toward  the  center,  L 
moves  away  from  it,  and  so  forth,  as  before. 

Parabola.  Corresponding  to  Theorem  9,  we  have  the  fol- 
lowing theorem,  the  proof  of  which  is  left  to 
the  student. 

THEOREM  10.  Let  PI  and  LI  be  pole  and 
polar  in  a  parabola,  and  let  the  diameter 
through  PI  meet  L^  in  P2  and  the  parabola  in 
Q.  Then  LI  is  parallel  to  the  tangent  at  Q, 
and  Q  is  the  mid-point  ofPiP.2. 
FIG.  28  Consequently,  if  a  point  P  traces  a  diameter 

*  Let  the  student  give  an  analytical  proof  of  these  facts  and  hence  of 
the  corollary  ;  cf.  Exs.  1,  2.  There  is  no  geometrical  proof,  analogous 
to  that  of  the  text.  The  r61es  of  Qi  and  Q2  in  that  proof  cannot  be 
played  here  by  the  points  in  which  D  meets  the  conjugate  hyperbola ; 
these  points  do  not  divide  P\Pi  harmonically:  LI  is  the  polar  of  PI 
with  respect  to  the  given  hyperbola,  and  not  with  respect  to  its  conjugate. 


DIAMETERS.     POLES  AND  POLARS  323 

D  of  a  parabola,  its  polar  L  moves  always  parallel  to  the 
tangent  at  the  point  in  which  D  meets  the  curve.  If  P  moves 
along  D  in  either  direction,  receding  indefinitely,  then  L  moves 
in  the  opposite  direction,  and  recedes  indefinitely. 

Theorems  9  and  10  furnish  new  methods  for  the  construc- 
tion of  the  polar  of  a  given  point  or  the  pole  of  a  given  line. 
These  we  shall  consider  in  the  next  paragraph. 

EXERCISES 

1.  Give  an  analytical  proof  of  the  second  part  of  Theorem 
9,  in  the  case  that  D  meets  the  conic. 

2.  The  same  if  the  conic  is  a  hyperbola  and  D  does  not  meet  it. 

3.  Theorem  9  is  no  longer  valid  if  the  conic  is  a  hyperbola 
and  P1  lies  on  an  asymptote.     Prove  that,  in  this  case,  L\  is 
parallel  to  the  asymptote,  and  that  the  product  of  the  distances 
OPi  and  OP2  is  constant,  where  P2  is  the  point  in  which  LI 
intersects  the  other  asymptote. 

4.  Prove  Theorem  10. 

5.  A  pair  of  conjugate  hyperbolas  and  a  point  P  are  given. 
Show  that  the  polars  of  P  with  respect  to  the  two  hyperbolas 
are  parallel  to,  and  equally  distant  from,  the  diameter  conju- 
gate to  the  diameter  through  P. 

By  applying  Th.  9,  the  Corollary,  and  Th.  10,  prove  the  fol- 
lowing theorems. 

6.  Let  C  be  a  hyperbola,  C'  the  conjugate  hyperbola,  and  D  a 
diameter  meeting  C'.    Then  the  polar  of  an  extremity  of  D  with 
respect  to  C  is  the  tangent  to  C'  at  the  other  extremity  of  D. 

7.  The  polar  of  a  point  with  respect  to  a  central  conic  (not 
a  circle)  is  perpendicular  to  the  line  joining  the  point  to  the 
center,  if  and  only  if  the  point  is  on  an  axis  of  the  conic. 

8.  If  a  line  is  normal  to  a  parabola  at  one  extremity  of  the 
latus  rectum,  its  pole  lies  on  the  diameter  passing  through  the 
other  extremity. 


324  ANALYTIC   GEOMETRY 

13.  Construction  Problems.  In  problems  of  construction  re- 
lating to  conies,  diameters  play  an  important  role.  To  con- 
struct a  diameter  of  a  conic,  one  has  but  to  draw  two  parallel 
chords  and  join  their  mid-points. 

Center,  Axes,  Foci.  Consider  first  a  central  conic,  drawn  on 
paper.  Construct  two  diameters ;  their  point  of  intersection 
will  be  the  center,  0,  of  the  conic.  With  0  as  center,  describe 
a  circle  cutting  the  conic  in  four  points ;  the  lines  through  0 
parallel  to  the  sides  of  the  rectangle  determined  by  the  four 
points  will  be  the  axes. 

If  the  conic  is  an  ellipse,  the  lengths  a  and  b  are  now  known, 
and  c  =  Va2  —  62  can  be  constructed  by  means  of  a  right  tri- 
angle ;  cf .  Ch.  VII,  Fig.  2.  Thus  the  foci  will  be  located. 

If  the  conic  is  a  hyperbola,  only  a  is  known.  But  then  6 
can  be  found  by  reversing  the  construction  of  Ch.  VIII,  §  9. 
Hence  the  asymptotes  and  foci  may  be  accurately  constructed. 

Let  a  parabola  be  given.  Construct  a  diameter  and  two 
chords  perpendicular  to  it;  the  line  joining  the  mid-points  of 
these  chords  is  the  axis  of  the  parabola.  The  construction  of 
the  focus  we  postpone  until  we  have  given  that  of  a  tangent. 

Tangents.  To  construct  the  tangent  to  a  central  conic  at  a 
point  P,  construct  the  center  0,  and  then  draw  OP  and  a  chord 
parallel  to  OP.  Let  K  be  the  mid-point  of  this  chord.  Then 
the  line  through  P  parallel  to  OK  is  the  tangent  at  P.  Why  ? 
If  the  conic  is  a  parabola,  construct  the 
axis.  LetTif  be  the  foot  of  the  perpendicular 
dropped  from  P  on  the  axis,  and  make  OM 
equal  to  OK  (Fig.  29).  Then,  by  Ch.  VI,  §  3, 
Ex.  8,  MP  is  the  tangent  at  P.  The  focus  can 
now  be  constructed  by  use  of  the  focal  prop- 
erty, namely,  by  constructing  the  focal  radius 
FIG.  29  ^  PF  as  the  line  making  £  MPF  =  %.  TP8. 

Of  course,  if  the  focus,  or  foci,  of  a  conic 
are  given,  the  tangent  at  a  point  can  be  constructed  by  means 
of  the  focal  property  of  the  conic. 


DIAMETERS.     POLES  AND  POLARS  325 

To  construct  the  tangents  to  a  conic  from  an  external  point 
is  a  more  difficult  problem.  We  shall  give  presently  a  solu- 
tion involving  poles  and  polars  and  refer  the  student  else- 
where for  one  based  on  more  elementary,  though  less  elegant, 
principles.* 

Poles  and  Polars.  Given  an  elementary  construction  for  the 
tangents  from  an  external  point,  we  can  carry  through  ac- 
curately the  constructions  of  §  %11  for  the  polar  of  a  given  point 
and  the  pole  of  a  given  line. 

We  are  more  interested,  however,  in  the  constructions  of 
poles  and  polars,  based  on  the  theorems  of  §  12.  We  will 
describe,  for  example,  the  construction  by  this  method  of  the 
polar  of  a  given  point  PI  with  respect  to  a  central  conic. 
Draw  the  diameter  D  through  Pl  (Fig.  26)  and,  by  drawing  a 
chord  parallel  to  D  and  bisecting  it,  construct  the  diameter  D' 
conjugate  to  D.  On  a  separate  sheet  construct  the  third  pro- 
portional to  the  length  OPt  and  the  half  length,  d,  of  ZXf 
Lay  off  the  resulting  length  from  0  on  D  in  the  proper  direc- 
tion, according  to  the  Corollary  of  Theorem  9,  and  through  the 
point  thus  reached  draw  the  line  parallel  to  D'.  This  line  is 
the  polar  of  P^ 

The  construction  is  the  same  whether  Pl  lies  inside,  on,  or 
outside  the  conic. 

Tangents  from  an  External  Point.  Let  the  point  be  P  and 
construct  its  polar  L  by  the  method  just  described.  The  lines 
joining  P  to  the  points  of  intersection  of  L  with  the  conic  are, 
by  §  11,  Th.  8&,  the  required  tangents. 

*  Cf.,  for  example,  Wentwortlrs  Plane  and  Solid  Geometry,  Ed.  of 
1900,  pp.  416,  435,  455. 

t  If  D  does  not  meet  the  conic  (as  may  happen  in  the  case  of  a  hyper- 
bola) and  the  conjugate  hyperbola  is  not  given,  the  length  d  (Fig.  27) 
is  unknown,  so  that  a  separate  construction  for  it  is  necessary.  Here  d 
equals  the  61  of  the  formula,  a^  —  b^  =  a2  —  62,  of  §  6,  Th.  10.  The 
lengths  a,  ft,  and  ai  are  known  or  can  be  constructed  by  methods  already 
given  ;  the  length  k  =  Va2  —  62  is  found  by  using  a  right  triangle  and, 
finally,  that  of  61  =  Va!2  —  fc2,  in  the  same  way. 


326  ANALYTIC   GEOMETRY 

EXERCISES 

1.  Using  a  templet,  draw  an  ellipse.     Carry  through  in  de- 
tail the  constructions  for  (a)  the  center,  (6)  the  axes,  (c)  the 
foci.     Devise  a  method  for  constructing  the  directrices. 

2.  The   same  problem   for  the  hyperbola.     Construct  also 
the  asymptotes. 

3.  Construct  the  axis,  a  tangent,  the  focus,  and  the  directrix 
of  a  parabola.  , 

4.  Construct  the  tangent  to  a  hyperbola  at  a  given  point  by 
use  of  the  focal  property.     Use  a  templet  to  draw  the  hyper- 
bola and  consider  that  the  foci  are  given. 

5.  The  same  for  an  ellipse. 

6.  Perform  in  detail  the  construction,  based  on  Theorem  9, 
§  12,  of  the  pole  of  a  given  line  with  respect  to  a  central  conic. 

7.  Carry   through    carefully    the    construction,    based   on 
Theorem  10,  §  12,  of  the  polar  of  a  given  point  with  respect 
to  a  parabola. 

8.  The  same  for  the  pole  of  a  given  line. 

EXERCISES  ON  CHAPTER  XIV 
DIAMETERS 

1.  Prove   that  two   similar  ellipses  with  the  same  center 
and  the  same  transverse  axis  have  the  same  pairs  of  conjugate 
diameters. 

2.  A  line   meets  a  hyperbola  in  the  points  Pt  and  P2  an(i 
meets  the  asymptotes  in  the  points  Ql  and  Q2.     Prove  that 
the  segments  PiP2  and  QiQ%  have  the  same  mid-points. 

3.  Using  the  result  of  Ex.  2,  show  that  any  two  hyperbolas 
with  the  same  asymptotes  have  the  same  pairs  of  conjugate 
diameters. 

4.  Prove  that  the  line-segment  joining  two  extremities  of 
conjugate  diameters  of  a  hyperbola  is  parallel  to  one  asym- 
ptote and  is  bisected  by  the  other. 


DIAMETERS.     POLES  AND  POLARS  327 

5.  The  chords  of  an  ellipse  from  a  vertex  to  the  extremi- 
ties of  the  minor  axis  are   parallel  to  a   pair  of   conjugate 
diameters.     Prove  this  theorem. 

6.  Two  chords  connecting  a  point  of  a  central  conic  with 
the  ends  of  a  diameter  are  called  supplemental  chords.     Show 
that  chords  of  this  nature  are  always  parallel  to  a  pair  of  con- 
jugate diameters. 

7.  Show  that,  if  a  parallelogram  has  its  vertices  on  a  cen- 
tral conic,  its  center  is  at  the  center  of  the  conic  ;  hence  prove, 
by  Ex.  6,  that  the  sides  of  the  parallelogram  are  parallel  to  a 
pair  of  conjugate  diameters. 

8.  Prove  that  the  angle  which  a  diameter  of  an  ellipse, 
not  an  axis,  subtends  at  a  vertex  is  the  supplement  of  the 
angle  which  the  conjugate  diameter  subtends  at  an  extremity 
of  the  minor  axis. 

9.  A  parallelogram  is  circumscribed  about  an  ellipse  by 
drawing  the  tangents  at  the  ends  of  a  pair  of  conjugate  diam- 
eters.    Prove  that  the  area  of  this  parallelogram  is  the  same, 
no  matter  what  pair  of  conjugate  diameters  is  chosen. 

Suggestion.     Compute  the  area   of   the  triangle  with   one 
diameter  as  base  and  an  extremity  of  the  other  as  vertex. 

10.  State   and   prove   the   corresponding   theorem   for   the 
hyperbola. 

11.  Show  that  in  the  case  of  the  hyperbola*  the  parallelo- 
gram of  the  two  preceding  exercises  always  has  its  vertices  on 
the  asymptotes. 

12.  Prove  that  the  segment  of  a  tangent  to  a  hyperbola  cut 
off  by  the  asymptotes  is  equal  in  length  to  the  diameter  paral- 
lel to  it. 

13.  Prove  that  the  tangents  to  a  central  conic  at  the  extrem- 
ities of  a  chord  meet  on  the  diameter  bisecting  the  chord. 

14.  Show  that  a  line  through  a  focus  of  a  central  conic  per- 
pendicular to  a  diameter  meets  the  conjugate  diameter  on  a 
directrix. 


328  ANALYTIC   GEOMETRY 

15.  Prove  that,  if  P  and  P'  are  extremities  of  a  pair  of  con- 
jugate diameters  of  a  central  conic,  the  normals  at  P  and  P' 
and  the  line  through  the  center  perpendicular  to  PP1  meet  in 
a  point. 

POLES  AND  POLARS 

16.  Find  the  polar  of  a  focus  of  a  central  conic  with  respect 
to  the  auxiliary  circle. 

17.  Prove,  for  a  central  conic,  that  the  line-segment  joining 
any  point  to  the  intersection  of  the  polar  of  the  point  with  a 
directrix  subtends  a  right  angle  at  the  corresponding  focus. 

18.  The  same  for  a  parabola. 

19.  Show,  for  a  central  conic,  that  any  chord  through  a  focus 
is  perpendicular  to  the  line  joining  the  focus  to  the  pole  of 
the  chord. 

20.  The  same  for  a  parabola. 

21.  Two  rectangular  hyperbolas  are  so   situated   that   the 
axes  of  one  are  the  asymptotes  of  the  other.     Prove  that  the 
polars  of  a  point  with  respect  to  the  two  hyperbolas  are  always 
perpendicular. 

22.  The  perpendicular  from  a  point  P  on  the  polar  of  P 
with  respect  to  a  central  conic  meets  the  transverse  axis  in  A 
and  the  conjugate  axis  in  jB.     Show  that  PA  :  PB  =  62 :  a2. 

23.  The  segment  of  the  axis  of  a  parabola  intercepted  by 
the  polars  of  two  points  is  equal  to  the  projection  on  the  axis 
of  the  line-segment  joining  the  two  points. 

Locus  PROBLEMS 

24.  A  line  is  drawn  through  the  focus  of  a  central  conic  per- 
pendicular to  a  variable  diameter.     Find  the  locus  of  the  point 
in  which  it  intersects  the  conjugate  diameter. 

25.  A  point  moves  so  that  its  polar  with  respect  to  an  ellipse 
forms  a  triangle  of  constant  area  with  the  axes  of  the  ellipse. 
What  is  its  locus  ? 


DIAMETERS.     POLES  AND  POLARS  329 

Ans.   A  pair  of  conjugate  rectangular  hyperbolas  with  the 
axes  of  the  ellipse  as  asymptotes. 

26.  Find  the  locus  of  the  poles,  with  respect  to  a  central 
conic,  of  the  tangents  to  a  circle  whose  center  is  the  center  of 
the  conic. 

27.  Find  the  locus  of  the  poles,  with  respect  to  the  circle 
X2  _j_  yz  —  a2?  of  t^  tangents  to  the  parabola  y1  =  2  mx. 

28.  Find  the  locus  of  the  poles,  with  respect  to  the  parabola 
y2  =  2  mx,  of  the  tangents  to  the  parabola  y2  =  —  2  mx. 

29.  Find  the  locus  of  the  mid-point  of  a  chord  of  an  ellipse, 
if  the  pole  of  the  chord  traces  the  auxiliary  circle. 

30.  The  same  for  a  hyperbola. 


CHAPTER   XV 


FIG.  1 


TRANSFORMATIONS  OF  THE   PLANE.     STRAIN 

1.  Translations.  Definition.  By  a  translation  of  a  plane 
region  S  is  meant  a  displacement  of  S  whereby  each  point  of 
S  is  carried  in  a  given  (fixed)  direc- 
tion by  one  and  the  same  given  dis- 
tance. Thus,  when  a  window  is 
raised,  a  pane  of  glass  in  the  window 
experiences  a  translation. 

It  is  not  important  what  particular 
region  S  is  considered.  Indeed,  it  is  usually  desirable  to 
consider  the  whole  unbounded  plane  as  S.  The  essential 
thing  is  the  above  law  which  connects  the  initial  position  of 
an  arbitrary  point  of  S  with  its  final  position. 

Analytic  Representation.  Let  P :  (x,  y}  be  an  arbitrary 
point  of  the  plane,  and  let  P' :  (xf,  y')  be  the  point  into  which 
P  is  carried  by  the  translation.  Let  a 
and  6  be  respectively  the  projections  of 
the  directed  line-segment  PP'  on  the 
axes  of  x  and  y.  Then 

(i)         {$!",+*      .•     • 

These  formulas  are  the  same  as  those  which  represent  a 
transformation  of  coordinates,  the  new  axes  being  parallel  to 
the  old  and  having  the  same  respective  directions.  But  the 
interpretation  of  the  formulas  is  wholly  different.  There,  the 
point  P  remained  unchanged.  It  had  new  coordinates  as- 
signed to  it  by  referring  it  to  a  new  set  of  axes.  Here,  the 

330 


TRANSFORMATIONS   OF   THE   PLANE 


331 


axes  do  not  change.  It  is  the  point  P  that  changes.  The 
point  P  is  picked  up  and  set  down  in  a  new  place,  namely, 
atP'. 

Example  1.     Represent  analytically  the  translation  whereby 
the  plane  is  carried  in  the  direction 
of  the  positive  avaxis  a  distance  of  2 
units : 

Solution  :        x'  —  x  +  2,         y'  =  y. 


Example  2.     Let  the  curve  C:- 


FIG.  3 


(2)  y  =  x*-x  +  % 

be  carried  in  the  direction  of  the  negative  axis  of  y  a  distance 
of  •§•  units.     What  will  be  the  equation  of  the  new  curve,  G'  ? 
The  formulas  representing  the  translation  are  : 


Hence        x  =  x',  y  =  y'  + 

and  equation  (2)  goes  over  into 


or 


y'  =  x'' 

FlG  4  The  new  curve,  C',  is  evidently  symmetric 

in  the  origin.  But  the  shape  of  C"  is  the 
same  as  the  shape  of  C.  Hence  O  is  symmetric  in  the  point 
A  :  (0,  f),  which  corresponds  to  the  origin. 

Example  3.  A  freight  train  is  running  northwest  at  the 
rate  of  30  miles  an  hour.  If  (x,  y)  are,  at  noon,  the  coordi- 
nates of  an  arbitrary  point  of  the  floor  of  one  of  the  platform 
cars,  referred  to  axes  directed  east  and  north  respectively,  de- 
termine the  coordinates  (x',  y')  of  the  same  point  t  hours 
later. 

Here,  the  components  of  PP',  after  one  hour  has  elapsed, 
are  clearly : 

a  =  30  cos  135°  =  -  15  V2,        b  =  30  sin  135°  =  15  V2. 


332  ANALYTIC   GEOMETRY 

After  t  hours  they  are 

a  =  -15V2f,        6  =  15V2*. 
Hence 

xf  =  x-  15  V2J,        y'=  y  +  15  V2 «. 

EXERCISES 

1.  Express  analytically  the  translation  which   carries   the 
origin  into  the  point  (2,  —  1),  and  hence  show  that  the  point 
(— 1,  2)  is  carried  into  (1,  1).     Draw  a  figure  showing  what 
happens  to  the  unit  circle,  xz  -f  yz  =  1. 

2.  Apply  the  translation  of  Ex.  1  to  the  curve 

2/  =  2«2  +  8a;  +  9. 

3.  Determine  a  translation  which  will  carry  the  curve 

y  =  4a;2  —  8a  +  3 
into  a  parabola  whose  equation  is  in  a  normal  form. 

4.  An  aeroplane  is  flying  at  the  rate  of  120  miles  an  hour 
on  a  straight,  horizontal  course  having  a  direction  30°  south  of 
east.     If  (x,  y)  are,  at  a  given  instant,  its  coordinates,  referred 
to  axes  directed  east  and  north  respectively,  determine  its  co- 
ordinates (x',  y')  after  t  minutes  have  elapsed. 

Prove  analytically  (i.e.  by  means  of  the  representation  (1) 
of  the  text)  the  following  theorems. 

5.  A  translation  carries  a  straight  line,  in  general,  into  a 
parallel  straight  line.     What  are  the  exceptions  ? 

6.  A  translation  carries  a  circle  into  a  circle  of  the  same 
radius. 

7.  A  translation  carries  two  mutually  perpendicular  right 
lines  into  two  mutually  perpendicular  right  lines. 

2.  Rotations.  Lefr  the  plane  be  rotated  about  the  origin 
through  an  angle  0.  What  will  be  the  coordinates,  (x1,  y'),  of 
the  point  P'  into  which  a  given  point  P,  with  the  coordinates 
(x,  y),  is  carried  ? 


TRANSFORMATIONS   OF  THE   PLANE 


333 


The  solution  can  be  read  off  at  sight  from  the  figure.     We 
have: 


(1) 


P:(x,V) 


Fia.  5 


x  =  UM  = 

x'=  OM',  y'  =  M'P'. 

Now,  Proj  OP'  =  Proj  OJtfi+  Proj  M^P', 

and  if  we  take  the  projections  first  along  the  axis  of  x,  and 
then  along  the  axis  of  y,  we  obtain  p'-(x'y') 

immediately  the  desired  relations  : 

f  x'  =  x  cos  0— y  sin  0, 
\  y'  =  x  sin  Q  +  y  cos  0. 

It  is  easy  to  solve  these  equa- 
tions algebraically  for  x  and  y ;  or  ^^ 
the  formulas  for  x  and  y,  in  terms 
of  x'  and  y',  can  be  written  down 
directly  by  projecting  the  broken  line  OM'P'  along  OMt  and 
perpendicularly  to  OJ/i : 

,o\  I  x=      x'  cos  0  4-  y' sin 6, 

\  y  =  —  x'  sin  0  +  y'  cos  0. 

Example.  It  is  clear  geometrically  that  a  circle  with  its 
center  at  the  origin  must  be  carried  over  into  itself  by  any  of 
the  above  rotations.  Let  us  see  what  the  analytic  effect  on 
its  equation  is  if  such  a  rotation  is  performed. 

The  equation  of  the  given  circle  is 

&  +  y*  =  p2. 

Replacing  x  and  y  by  their  values  from  (2),  we  have : 
(xf  cos  6  +  y'  sin  0)2  +  (—  x'  sin  &  +  y'  cos  0)2=  p2, 


or 


COS20 

+  sin2  0 


a'2  +  2  sin  0  cos  0 
—  2  sin  0  cos  6 


x'y'  +  sin2  0 
+  cos2  0 


Hence  x1*  +  y'2  .=  p '-, 

and  we  get  the  same  circle,  as  we  should. 


334  ANALYTIC  GEOMETRY 

EXERCISES 

1.  Write  down  directly  from  a  figure  the  formulas  which 
represent  a  rotation  of  90°  about  the  origin,  and  verify  the  re- 
sult by  substituting  0  =  90°  in  (1). 

2.  Show   that,  if    the    curve   xy  —  2  a?   is    rotated    about 
the   origin    through    an    angle    6  =  —  45°,   its   equation   goes 
over  into  the  usual  form  of  the  equation  of  an  equilateral 
hyperbola. 

3.  Kotate  the  parabola  y*  =  2mx  through  —  90°  about  the 
origin. 

4.  Prove  analytically  that,  if  an  arbitrary  straight  line  be 
rotated  about  the  origin  through  90°,  the  new  line  will  be  per- 
pendicular to  the  old  one. 

5.  Prove  that,  if  an   arbitrary  line  be  rotated  about   the 
origin  through  the  angle  6,  the  angle  from  this  line  to  the  new 
line  will  be  0. 

3.  Transformations  of  Similitude.  Let  the  plane  be  stretched, 
like  an  elastic  membrane,  uniformly  in  all  directions  away  from 

the  origin.  This  tranformation 
is  evidently  represented  analyti- 
cally by  the  equations  : 

/-i\ 

where   k    is   a   constant   greater 
FIG  6  than  unity.     If  k  is  positive,  but 

less  than  unity,  the  transforma- 
tion represents  a  shrinking  toward  the  origin.  The  stretchings 
and  shrinkings  defined  by  (1)  are  known  as  transformations  of 
similitude. 

These  transformations,  like  the  translations  and  the  rota- 
tions, preserve  the  shapes  of  all  figures ;  but,  unlike  those 
transformations,  they  alter  the  sizes  of  figures. 


TRANSFORMATIONS   OF  THE   PLANE  335 

Example.     The  equilateral  hyperbola 
y?  —  f  =  a2 

is  carried  by  (1),  if  Jc  is  taken  equal  to  -: 

a 

»'  =  -,   y' =  %-,        or        x=ax',   y  =  ayf, 
a  a 

into  the  curve 

aV2  -  a2?/'2  =  a2, 

or  a/2  —  y'2  =  1. 

Thus  all  equilateral  hyperbolas  are  seen  to  be  similar  to  one 
another,  since  each  can  be  transformed  by  (1)  into  the  particu- 
lar equilateral  hyperbola 


Inverse  of  a  Transformation.     The  transformation, 

x' 


(2) 


nfl   

*  — 


obtained  by  solving  the  formulas  (1)  for  x,  y,  is  called  the  in- 
verse of  the  transformation  (1).  In  general,  if  a  given  trans- 
formation carries  (x,  y)  into  (x'}  y'),  the  transformation  carry- 
ing (a/,  y')  into  (x,  y)  is  known  as  the  inverse  of  the  given 
transformation.  Thus,  the  rotation  (2),  §  2,  is  the  inverse  of 
the  rotation  (1),  §  2. 

It  is  clear  that  the  effect  of  the  inverse  transformation,  if 
performed  after  the  given  one,  is  to  nullify  the  given  one. 
Thus  (1),  §  2,  rotates  all  figures  through  the  angle  0,  and  then 
(2),  §  2,  rotates  them  through  the  angle  —  0,  i.e.  back  into 
their  original  positions. 

EXERCISES 

1.  Show  that  the  parabola  y2  =  2  mx,  0  <  m,  can  be  trans- 
formed by  (1)  into  the  parabola  t/2  =  x.  What  value  must  be 
taken  for  Jc  ? 


336 


ANALYTIC   GEOMETRY 


2.    Show   that  the  effect  of  performing  transformation  (1) 
and  then  transformation  (2)  is  to  leave  the  plane  unchanged. 

4.  Reflections  in  the  Axes.  Let  the  plane  be  reflected  in 
the  axis  of  x.  In  other  words,  let  it  be  rotated  through  180° 
about  the  axis  of  x.  Let  P:  (x,  y)  be  an 
arbitrary  point,  and  let  P ' :  (x',  y')  be  the 
point  into  which  P  is  carried.  Then, 
obviously, 


(1) 


x'  =x 


FIG.  7 


Similarly,  a  reflection  in  the  axis  of  y  is 
represented  by  the  formulas  : 

f  x'  =  —  x 

v         I,:*  • 


The  condition  that  a  curve  be  symmetric  in  one  of  the  axes 
(cf.  Ch.  V,  §  2)  is  obtained  at  once  from  these  transformations. 
Thus  the  curve  C  will  be  symmetric  in  the  axis  of  x  if  the 
curve  C1,  into  which  C  is  carried  by  (1),  is  the  same  curve  as  C; 
and  the  test  for  this  is,  that  the  equation  of  C  be  essentially 
unchanged  when  the  transformation  (1)  is  performed  on  it. 

For  example,  if  C  is  the  curve 

7/4  +  a;2  =  2  y2  +  y?, 

its  equation  is  unchanged  by  (1),  and  hence  C  is  symmetric  in 
the  axis  of  x.  But  it  is  changed  by  (2),  and  C  is,  therefore, 
not  symmetric  in  the  axis  of  y. 

Isogonal  Transformations.  A  transformation  is  said  to  be 
equiangular  or  isogonal  if  the  angle  which  any  two  intersecting 
curves,  Ci  and  G>,  make  with  each  other  is  the  same  as  the 
angle  which  the  transformed  curves,  C\  and  C"2,  make  with 
each  other. 

All  of  the  transformations  considered  thus  far  are  evidently 
isogonal.  We  turn  now  to  a  transformation  which  is  not. 


TRANSFORMATIONS   OF   THE   PLANE 


337 


EXERCISE 

Show  that  the  equations  of  the  inverse  of  the  reflection  (1) 
[or  (2)]  are  precisely  of  the  same  form  as  the  equations  of  the 
reflection.  A  transformation  for  which  this  is  true  is  said  to 
be  involutory. 

5.  Simple  Elongations  and  Compressions.  Let  the  plane  be 
stretched  directly  away  from  the  axis  of  x,  so  that  each  point 
is  carried,  along  a  parallel  to  the 
axis  of  y,  to  twice  its  original  dis- 
tance from  the  axis  of  x  (Fig.  8). 
Evidently,  the  analytic  condition 
is  that 

x'  =  x,  y'  =  2y- 

More     generally,     if     a     point 
P :  (x,  y)  is  to  be  carried  to  I  times 
its  original  distance  from  the  axis 
of  x,  where  I  may  have  any  positive  constant  value,  not  unity, 
the  transformation  will  be  given  by  the  formulas : 


FIG.  8 


(1) 


X  =  X, 

y'  =  iy- 

When  I  is  greater  than  unity,  these  formulas  represent  an 
elongation  $  when  I  is  less  than  unity,  they  represent  a  com- 
pression. 

If  the  elongation  is  away  from  the  axis  of  y  or  the  compres- 
sion is  toward  it,  then 
(2)  \*  =  *x, 

where  k  is  greater  than  unity  in  the  first  case  and  less  than 
unity,  but  positive,  in  the  second. 

These  transformations  were  discussed  geometrically  in  Ch. 
XIV,  §  7.  There  we  called  them  one-dimensional,  or  simple, 
elongations  and  compressions;  or,  jointly,  one-dimensional 
strains. 


338  ANALYTIC   GEOMETRY 

Example  1.     Let  the  circle 
i)  x*  +  y*=l 

be  subjected  to  the  transformation  (1).     Then  it  goes  over  into 

*"+f=i. 

Thus  the  circle  i)  is  carried  into  the  ellipse 
..,  cc2  ,  y1     .,  ,       7 

11)  T+&=1'  b  =  l> 

whose  axis  lying  along  the  axis  of  x  is  identical  with  the  cor- 
responding diameter  of  the  circle,  but  whose  axis  lying  along 
the  axis  of  y  is  the  corresponding  diameter  of  the  circle 
stretched  in  the  ratio  1  :  1. 

Example  2.     Let  the  ellipse  ii)  be  subjected  to  the  trans- 
formation (2).     Then 

*  +  £=! 

fc2          ft2 

Thus  the  ellipse  ii)  is  carried  into  the  ellipse 


From  these  examples  we  see  that  the  particular  circle  i)  can  be 
carried  by  means  of  two  one-dimensional  strains  into  an  arbi- 
trary ellipse  iii)  whose  axes  lie  along  the  axes  of  coordinates. 

Exercise.  Show  that  the  circle  i)  can  be  carried  into  the 
ellipse  iii)  by  a  single  one-dimensional  strain  and  a  transforma- 
tion of  similitude. 

Product  of  Two  Transformations.     The  combined  effect  of 
the  two  transformations  of  Examples  1  and  2  can  be  repre- 
sented analytically  as  follows.     First,  we  have 
(a)  x'  =  x.  y'  —  ly- 

Next,  the  point  (x',  y'}  is  carried  into  (a;",  y")  by  the  trans- 
formation 
(6)  x"  =  kx',          y"  =  y'. 

Eliminating  the  intermediate  stage  (x',  y'~),  we  get  : 
(c)  x"  =  Jcx,  y"  =  ly. 


TRANSFORMATIONS   OF  THE   PLANE 


339 


A  transformation,  as  (c),  which  arises  as  the  result  of  two 
successive  transformations,  as  (a)  and  (&),  is  called  the  product  of 
these  transformations.  Similarly,  (a)  and  (6)  are  spoken  of  as 
the  factors  of  (c) ;  or  (c)  is  said 'to  be  factor  ed  into  (a)  and  (6). 

Let  the  student  verify  the  fact  that,  if  the  circle  i)  is  sub- 
jected to  the  transformation  (c),  it  is  carried  over  into  the 
same  ellipse  iii)  into  which  i)  was  carried  by  the  successive 
applications  of  the  transformations  (a)  and  (6). 

Properties  of  the  Transformation.  One  of  the  most  impor- 
tant properties  of  one-dimensional  strains  is  that,  like  the  trans- 
formations previously  studied,  they  carry  straight  lines  over  into 
straight  lines. 

This  was  proved  geometrically  on  p.  304.     The  transforma- 
tion considered  there   is  given  analytically  by  (1).     It  was 
proved  also  that,  if  L  is  a  line  of  slope  A,  the  slope  of  the  line 
into  which  L  is  carried  by  (1)  is  *  „ 
(3)  A'  =  ZA. 

From  the  theorem  contained  in  formula  (3),  it  is  seen  that 
a  one-dimensional  strain  carries  parallel  lines  into  parallel  lines. 

Consider  an  arbitrary  curve,  C.     Its 
slope  at  any  one  of  its  points,  P,  is  v 


Perform  the  transformation  (1)  on  C. 
Then  PM  remains  unchanged  in 
length  ;  but  MQ  goes  over  into 


Hence  the  slope,  A',  of  C"  is 


FIG.  9 


q'=p'P'M'      Q±P  PM 
or  A'  =  ZA. 

We  have  thus  extended  the  validity  of  formulas  (3). 

*  The  proofs  on  p.  304  were  given  for  compressions,  but  they  are  valid, 
also,  for  elongations. 


340 


ANALYTIC  GEOMETRY 


It  follows  from  this  extension  that,  if  two  curves,  C\  and  (72, 
are  tangent  to  each  other,  the  transformed  curves,  G\  and  (72, 
will  also  be  tangent.  Fpr,  if  d  and  (72  are  tangent,  they  have 
the  same  slope  X  at  the  point  of  tangency.  Hence,  at  the  cor- 
responding point,  C"i  and  C"2  will  each  have  the  slope  A'  =  l\ 
and  consequently  will  be  tangent  to  each  other. 

Angles  are  not  in  general  preserved  by  a  one-dimensional 
strain.  It  is  true  that  right  angles  whose  sides  are  parallel 
respectively  to  the  coordinate  axes  go  over  into  right  angles 
satisfying  the  same  condition.  But  consider,  for  example,  the 
angle  between  a  line  L  and  the  axis  of  a;  (p.  304,  Fig.  13). 
L  is  carried  into  L  by  the  transformation  (1),  and  the  axis  of  x 
remains  fixed.  It  is  clear,  then,  that  the  new  angle  is  not 
equal  to  the  original  one. 

The    areas    of    figures,    also,    are 
changed,    and    changed    in    precisely 
the  ratio  of  I  (or  fc)  :  1.     This  is  obvi- 
£  ously  true  for  rectangles  whose  sides 

/I  W  are   parallel   to  the   coordinate  axes. 

\J  Jr  The  area,  A,  of  any  other  figure  is  the 

limit  approached  by  the  sum,  JB,  of 
the  areas  of  rectangles  inscribed  as 
shown  in  the  drawing : 

A  =  lim  B. 

By  the  transformation  (1),  A  is  carried 
— x    into  A'  and  B  into  B' ;  evidently 

A'  =  lim  B'. 

Since  the  area  of  each  rectangle  represented  in  the  sum  B' 
is  I  times  the  area  of  the  rectangle  from  which  it  originated, 

B'  =  IB. 

Hence  A'  =  lim  IB  =  I  lim  B, 

or  A'  =  IA,  q.  e.  d. 

Example.  The  area  of  the  circle  i)  is  ir.  It  follows,  then, 
that  the  area  of  the  ellipse  ii)  is  TT  b.  Applying  the  method 


FIG.  10 


TRANSFORMATIONS   OF   THE   PLANE  341 

again  to  this  ellipse,  we  obtain  as  the  area  of  the  ellipse  iii) 
the  value  -n-  ab.     We  have  thus  obtained  the  following  result. 

The  area  of  the  ellipse 

££+£  =  1 

a2     bz 
is  A  =  irab. 

EXERCISES 

1  .   Show  that  the  circle  #2  +  y*  =  a2  can  be  carried  into  the 
ellipse 


by  a  one-dimensional  strain  ;  cf.  p.  306. 

2.  Prove  that  the  rectangular  hyperbola  a?  —  y1  =  a2  can  be 
carried  into  the  hyperbola 

^_£=1 

a2     62 

by  a  one-dimensional  strain. 

3.  In  Examples  1,  "2  of  p.  338  can  the  order  of  the  trans- 
formations be  reversed  ?     Prove  your  answer. 

4.  A  one-dimensional  strain  changes,  in  general,  the  shapes 
of  curves.     Is  this  true  in  all  cases?     For  example,  in  the 
case  of  a  parabola  ? 

5.  Prove  analytically  that  the  transformation  (1)  carries  a 
straight  line  into  a  straight  line. 

6.  Show  analytically  that,  if  a  line  L  is  carried  by  (1)  into 
a  line  L',  the  slopes  of  L  and  L'  are  connected  by  formula  (3). 

7.  Find  the  equations  of  the  transformation  which  is  the 
product  of  the  two  transformations  : 

x'  =  x-l,  y'  =  y  +  2,        x"  =  x',  y"  =  -y'. 

8.  The  same  for  the  rotation  about  the  origin  through  45°, 
followed    by  the    translation  which    carries   the   origin   into 
the  point  (3,  —  1). 


342  ANALYTIC   GEOMETRY 

9.    The  same  for  the  transformation  of  similitude  which 
doubles  all  the  lengths,  followed  by  a  reflection  in  the  axis  of  y. 

10.  Prove  analytically,  for  the  following  transformations, 
that  the  product  of  a  transformation  and  its  inverse  is  the 
identical  transformation,  x'  =  x,  y'  =  y. 

(a)  translations  ;         (6)  rotations  ; 

(c)   reflections  ;  (d)  one-dimensional  strains. 

11.  Factor  the  transformation 

(a)  into  two  one-dimensional  strains ; 

(&)  into  a  one-dimensional  strain  and  a  transformation  of 
similitude. 

12.  Prove  that  the  rotation  about  the  origin  through  180°  — 
also  called  the  reflection  in  the  origin  —  is  the  product  of  the 
reflections  in  the  axes. 

13.  Factor  the  transformation 

x'  =  4#,  y'  =  —  2y 

into  two  one-dimensional  strains  and  the  reflection  in  the  axis 
of  x.     How  else  can  it  be  factored  ? 

Express  each  of  the  following  transformations  as  the  prod- 
uct of  two  or  more  simple  transformations. 

14.  x  =  —  3x,  y'  =  —  2y. 

15.  x'  =  —  x  +  2,  y'  =  y  —  3. 

16.  X'  : 


17.  a/  =  ^=^-6,         2/'  =  ^±^  +  3. 

V2  V2 

6.   The  General  Affine  Transformation.      By  the    title    is 
meant  the  transformation 

|  x'  =  ax+  by  +  c, 

where  A  =  ab'  —  a'b  =£  0. 


TRANSFORMATIONS  OF  THE   PLANE  343 

All  of  the  foregoing  transformations  come  under  this  type, 
but    there    are    transformations    comprised    under    (1),   for 
example, 
(2)  x'  =  2x-3y  +  l,        y'  =  _aj  +  4y-2, 

which  are  not  of  any  of  the  above  forms.     We  shall  prove  the 
following  theorem. 

THEOREM  1.  The  transformation  of  the  plane  defined  by 
means  of  equations  (1)  can  be  generated  by  a  succession  of  the 
transformations  studied  in  §§  1-5.  In  other  words,  it  can  be 
FACTORED  itito  transformations  of  the  type  of  those  o/§§  1-5. 

Proof.  If  c  and  c'  are  not  both  0,  let  the  (x,  2/)-plane  be 
subjected  to  a  translation : 


where  £,  rj  are  arbitrary  and  shall  be  determined  presently. 
Thus  equations  (1)  are  replaced  by  the  following : 

\x'  =  ax!  +  byl—(a£  +  brj  —  c), 
\y'  =  a'x,+b'yl-(a'^b'ri-cf). 

We  now  determine  £  and  rj  so  that  both  parentheses  will 
disappear.  This  is  done  by  solving  the  simultaneous  linear 
equations : 

a£  +  brj  —  c  =  0, 
a'£  +  b'rj  -  c'  =  0. 

The  solution  is  always  possible  and  unique,  since,  by  hypothe- 
sis, A  =  abr  —  a'b  3=  0. 

We  thus  have  a  simpler  pair  of  equations  to  study,  namely, 

(5)  f  yf  =  a*i  +  tyi, 

It  will  be  sufficient,  then,  if  we  can  prove  our  theorem  for 
the  case  that  c  =  0  and  c'  =  0,  i.e.  for  the  pair  of  equations 

x' —  ax+  by, 
/'  =  a'x  +  b'y, 


344 


ANALYTIC   GEOMETRY 


for  we  have  just  seen  that  we  can  pass  from  equations  (1)  to 
equations  (5)  by  a  translation,  and  this  is  one  of  the  transfor- 
mations admitted  by  the  theorem. 

Consider  an  arbitrary  circle  with  its  center  at  the  origin : 

Let  us  see  into  what  curve  it  is  carried  by  (6). 

To  do  this,  solve  equations  (6)  for  x  and  y.     The  result  is 
the  inverse  of  (6),  namely : 

lx  =  Atf  +  By't 

V    /  A  t  ~*t     \      T?//ii/ 


where 


and 


'  =  --,    A'  =  --,    £'  =  -, 
A  A  .A 


Next,  substitute  these  values  for  x  and  y  in  (7) : 
(9)    (A*  +  A'*)  x">  -f  2(AB  +  A'B')  x'y'  +  (B2  +  B'2)  y'*  =  P\ 

The  locus  of  this  equation  is  an  ellipse  with  its  center  at 
the  origin.  For,  first,  the  equation  has  a  locus,  since  all  the 

points  (xf,  y'}  into  which  the  points 
(x,  y)  of  the  circle  (7)  are  carried  by 
(6)  lie  on  (9).  Secondly,  the  locus 
does  not  extend  to  infinity  in  any 
direction.* 

In  general,  this  ellipse  will  not 
be  a  circle.  Let  L'  be  the  extremity 
of  an  axis.  Since  the  transforma- 
tion (8)  carries  any  straight  line 
through  the  origin  into  a  straight 
line  through  the  origin,  OL'  will 
correspond  to  a  certain  radius  OL 
of  the  circle  (7). 

We   now  rotate   the  (x,  ?/)-plane 

*Or,  the  expression  &  —  4  A C  (Ch.  XII,  §3),  formed  for  (9),  is 
negative.: 


FIG.  11 


TRANSFORMATIONS   OF  THE   PLANE  345 

about  the  origin  through  an  angle  0  such  that  OL  comes  to  lie 
along  the  positive  axis  of  x : 

xl  =  x  cos  6  —  y  sin  0, 
yl  =  x  sin  6  +  y  cos  0. 

Furthermore,  we  rotate  the  (x',  t/')-plane  about  the  origin 
through  such  an  angle  6'  that  OL'  comes  to  lie  along  the 
positive  axis  of  x : 

i  =  x'  cos  6'  —  y'  sin  6', 


\  y/  =  x'  sin  0'  +  y'  cos  6'. 

What  is  the  final  result?  Obviously  the  following.  An 
arbitrary  point  (xlt  y^)  of  the  plane  is  carried  by  the  inverse  of 
(10)  into  a  point  (x,  y}  ;  this  point  is  carried  by  (6)  into  a 
point  (x',  yf)  ;  and  finally  the  point  (x',  y')  thus  obtained  is 
carried  by  (11)  into  (»/,?//).  To  write  out  these  transforma- 
tions explicitly  would  be  a  long  piece  of  work  ;  but  it  is  not 
necessary  to  do  so.  For,  first  of  all,  each  is  linear  and  leaves 
the  origin  unchanged.  Hence  the  final  transformation,  carry- 
ing the  point  (x^  y^)  directly  into  the  point  (#/,  y/),  is  also 
linear,*  and  it  leaves  the  origin  unchanged.  It  is,  then,  of  the 
form  : 


Consider  next  what  we  know  about  this  transformation. 

i)  It  carries  the  positive  axis  of  xv  over  into  itself.  Hence, 
when  yl  =  0,  y/  must  also  vanish,  no  matter  what  value  x±  may 
have;  thus 

Q  =  yxl, 

and,  consequently,  y  =  0. 

ii)  The  axes  of  the  ellipse  which  corresponds  to  the 
circle, 

(13)  *i2  +  2/12  =  P2, 

are  the  coordinate  axes. 

*  Cf  .  Ex.  1  at  the  end  of  the  chapter. 


346  ANALYTIC   GEOMETRY 

To  find  the  equation  of  the  ellipse,  solve  (12)  for  ajj,  y^  re- 
membering that  y  =  0 : 

=  21V  -1-  Sy/, 


aw, 


Thus  (13)  is  seen  to  go  into  the  ellipse 

(15)  21  V2  +  2  2l#*i'</i'  +  (S32  +  £>2)  y/2  =  P7- 

Since  the  axes  of  the  ellipse  (15)  lie  along  the  coordinate 
axes,  the  term  in  x^y^  must  disappear,  and  so  we  must  have 

2133  =  0. 
Now,  21  cannot  be  0,  for  A  =  I/a.     Hence  33  must  vanish  : 

33  =  0. 
It  follows,  then,  that 

0  =  0, 

and  thus  (12)  reduces  to  the  transformation: 


Moreover,  since  the  positive  axis  of  xl  goes  over  into  the 
positive  axis  of  #,,  we  see  that 

0<  «. 

But  8  may  be  negative.  In  this  case,  a  reflection  in  the  axis 
of  xl  will  change  the  sign  of  8,  and  hence  the  case  is  reduced 
to  the  one  in  which  8  is  positive  : 

0<8. 

Finally,  the  transformation  (16)  is  the  product  of  two  one- 
dimensional  strains  : 

x1  =  a  x,  I  x'  =  x, 


one  along  the  axis  of  x  and  one  along  the  axis  of  y. 

Let  us  now  recapitulate.     The  point  (x,  y}  is  carried  into  the 
point  (ajj,  2/j)  by  the  rotation  (10)  ;  (xl}  y^  is  carried  into  (#/,  y\) 


TRANSFORMATIONS   OF  THE   PLANE  347 

by  (16),  which  is  a  product  of  two  one-dimensional  strains  and, 
perhaps,  the  reflection  in  the  axis  of  x  ;  finally,  (x/,  y/)  is 
carried  into  (x',  y')  by  the  inverse  of  (11),  another  rotation. 
The  transformation  (6)  is,  then,  the  product  of  these  trans- 
formations in  the  order  enumerated.  The  proof  of  Theorem  I 
is  thus  complete. 

Properties  which  all  the  component  transformations  of  (1) 
have  in  common  are  also  properties  of  (1).  Consequently,  the 
general  qffine  transformation  carries  straight  lines  into  straight 
lines,  parallel  lines  into  parallel  lines,  and  tangent  curves  into  tan- 
gent curves.  It  does  not,  in  general,  preserve  angles  or  areas. 

Isogonal  Transformations.  Of  the  component  transformations 
of  (6),  the  rotations  (10)  and  (11)  always  preserve  angles. 
This  is  true  of  the  transformation  (16)  if  and  only  if  it  is  a 
transformation  of  similarity,  with  or  without  a  reflection  in  the 
axis  of  x,  i.e.  if  and  only  if  8  =  ±  a.  Hence  the  most  general 
isogonal  transformation  of  the  form  (6)  is  the  product  of  the 
rotation  (10),  the  transformation 


and  the  inverse  of  the  rotation  (11).     This  product  is  easily 
found  to  be  • 

x'  =    P[>cos(0:F0')-;ysin(0  qp0')], 
y'  =  ±p[>sin(0=F0')  +  y  cos(0  T  0')]. 

But  the  angle  6  —  6'  (or  Q  +  0')  is  no  more  general  than  a 
single  angle,  which  we  may  denote  by  <£.  Thus  the  result  can 
be  written  as 

x'  =  p(x  cos  <f>  —  y  sin  <£),         y'  =  ±  p(x  sin  <f>  -f  y  cos  <£). 

Replacing  x,  y  by  x1}  yl}  so  that  this  transformation  of  the 
form  (6)  reverts  to  the  form  (5),  and  then  applying  the  trans- 
lation (3),  we  obtain  as  the  most  general  transformation  of  the 
form  (1)  which  is  isogonal  : 

x'  =      p(x  cos  <£  —  y  sin  <£)  -f  c, 


(17) 

}  y'  =  ±  p(x  sin  <£  +  y  cos  </>)  +  c'. 

Here 


348  ANALYTIC   GEOMETRY 

(18)  a=pcos<j>,     b=—  /osin<£,     a'  =  ±  p  sin  <f>,     b'=±pcos<f>, 
and  hence, 

(19)  b'  =  ±a,  o'  =  T  6. 

Conversely,  every  transformation  (1),  for  which  (19)  is  true 
for  one  set  of  signs,  can  be  written  in  one  of  the  forms  (17) 
and  hence  is  isogonal.  For,  if  a,  b,  a',  b'  are  given,  satisfying 
(19)  for  one  set  of  signs,  values  of  p,  cos  <£,  and  sin<£  can  be 
found,  so  that  equations  (18)  hold  for  the  same  set  of  signs. 
These  values  are,  namely, 

,  -  a  —  b     % 

=Va2  +  &2,        cos  <ft  =     .  ,         sin  <j>  = 

' 


p=a  +     ,        cos  <    =     .  ,         sin  <>  =  —  . 

Va2  +  62'  Va2  +  W 

The  following  theorem  summaries  our  results. 

THEOREM  2.  TJie  transformation  (1)  is  isogonal  when  and 
only  when  either 

b'  =  a     and     a'  =  —  b         or         b'=  —  a     and     a'=b. 
If  it  is  isogonal,  it  can  be  written  in  one  of  the  forms  (17). 

Homogeneous  and  Non-Homogeneous  Transformations.  A 
polynomial  in  x  and  y  is  homogeneous,  if  its  terms  are  all  of 
the  same  degree  in  x  and  y*  Thus,  the  left-hand  sides  of  for- 
mulas (6)  are  homogeneous  polynomials  of  the  first  degree.  Ac- 
cordingly, a  transformation  of  the  form  (6)  is  called  a  homo- 
geneous affine  transformation  ;  and,  in  distinction,  a  transforma- 
tion of  the  form  (1),  where  c  and  c'  are  not  both  zero,  a  non- 
homogeneous  affine  transformation. 

Since  (1)  can  be  reduced  to  (6)  by  means  of  a  translation,  we 
have  the  theorem. 

THEOREM  3.  A  non-homogeneous  affine  transformation  is  the 
product  of  a  translation  and  the  corresponding  homogeneous 
transformation. 

*  Only  those  terms  with  non-vanishing  coefficients  need  be  considered, 
since  a  term  whose  coefficient  vanishes  has  the  value  of  0,  and  0  is  not  de- 
fined as  having  a  degree.  For  example,  if,  in  the  polynomial  x2  +  2  xy 
+  ax,  a  has  the  value  0,  the  polynomial  is  homogeneous  of  the  second 
degree. 


TRANSFORMATIONS   OF   THE   PLANE 


349 


EXERCISES 

1.  Prove   analytically    that   the   affine   transformation   (1) 
carries  a  straight  line  into  a  straight  line. 

2.  Show  that,  if  L  is  a  line  of  slope  A,  the  line  into  which  L 
is  carried  by  (1)  has  the  slope 

v=  a'+  b'X 
a+b\ 

3.  Using  the  result  of  Ex.  2,  prove  that  (1)  carries  parallel 
lines  into  parallel  lines. 

7.  Factorization  of  Particular  Transformations.  We  proceed 
to  illustrate  the  theory  of  the  preceding  section  by  carrying  it 
through,  step  by  step,  for  a  particular  case. 

Let  the  given  transformation  be 

(1)  •       x'  =  x  +  3  y,        y'  =  —  3  x  —  y. 

The  first  step  is  to  find  the  ellipse  into  which  the  circle, 

(2)  tf  +  ^  =  P», 

is  carried  by  (1).     Solving  equations  (1)  for  x  and  y,  and  sub- 
stituting the  values  obtained,  namely, 


in  (2),  we  have,  finally, 

(3)        5  x'2  +  6  x'y'  +  5  y"1  =  32  p2. 

This  ellipse  is  as  shown.  One  axis, 
OL',  lies  along  the  line  x'+y'  =  0. 
By  adding  equations  (1),  we  have 


and  hence  the  radius  OL  of  the  circle 
which  is  carried  into  OL'  is  along  the 
line  x  —  y  —  0.  Furthermore,  if  L' 
lies  in  the  fourth  quadrant,  as  in  the 
figure,  then  L  lies  in  the  first  —  not 
in  the  third. 


FIG>  12 


350  ANALYTIC   GEOMETRY 

Consequently,  a  rotation  of  the  (x,  y)-plane  about  0  through 
-45°: 

/A\ 

(4)       •  a 


V2  V2 

brings  OL  to  lie  along  the  positive  axis  of  x.     And  a  rotation 
of  the  (xf,  2/')-plane  about  0  through  +  45°  : 


r  '  -  x  — 
Xl  '  ~ 


does  the  same  for  OL'. 

The  next  step  is  to  find  the  transformation  carrying  fa,  3^) 
directly  into  fa',  #/).  This  we  do  by  eliminating  x,  y  and 
x',  y'  from  (1),  (4),  (5).  Thus, 


V2  V2 

or 

(6)  a51'  =  4s1,          yi'  =  2yi. 

Finally,  we  solve  (5)  for  x'  and  y'  : 

n\  x' 


V2  V2 

The  transformation  (1)  is  now  seen  to  be  the  product  of  the 
transformations  (4),  (6),  and  (7) ;  (4) 
carries  fa  y)  into  fa,  y^  by  a  rotation 
about  0  through  —  45° ;  (6)  carries 
fa,  yi)  into  fa',  y~i)  by  two  one-di- 
mensional strains  ;  (7)  carries  fa',  t//) 
into  (a;',  y')  by  another  rotation  about 
-*  0  through  -  45°. 

FIG.  13  Simplifications    in    Technique.      In- 

stead  of    seeking   the   ellipse   in   the 

(x',  y')-plane  into  which  the  circle  (2)  is  carried,  we  might 
equally  well  ask  for  the  ellipse  in  the  fa  ?/)-plane  which  is 


TRANSFORMATIONS  OF  THE  PLANE  351 

carried  into  the  circle 

(8)  ^-fy'2  =  p2. 

The  roles  of  the  two  planes  are  merely  reversed.  Adopting 
this  procedure  obviates  the  necessity  of  solving  (1)  for  the 
values  of  x,  y  to  be  substituted  in  (2).  For  now  (2)  is  replaced 
by  (8)  and  x'  and  y'  are  given  by  (1).  Thus  one  step  in  the 
process  is  eliminated.  The  others  remain  unchanged. 

Another  simplification  arises  in  factoring  a  transformation 
of  the  form 
/ON 


'  =  -3x-    y-2. 
Instead  of  proceeding  as  in  §  6,  we  set 

K  =  *  +  4>     and     /•- 

(y'  =  y-2,  [y  =  -3x-     y. 

Thus  (9)  is  the  product  of   the  transformation  (1)   and   the 
translation  which  carries  the  origin  into  the  point  (4,  —  2). 

EXERCISES 

Factor  the  following  transformations,  using  the  simplified 
method. 

1.  x'  =  x  +  3y,.  y'  =  -3x-y. 

2.  x'  =  3x-2y,  y'  =  -2x  +  3y. 

3.  x'  =  5x  +  lly,  y'  =  Wx+2y. 

4.  xt'=6x  +  18y  —  2,  y1  =  ITx  +  y  +  3. 

5.  «/sslllaj  +  4y  +  l,  y'  =  52x+78y-7. 


8.  Simple  Shears.  In  a  rectangle  with  its  center  at  the 
origin  and  with  its  sides  parallel  to  the  coordinate  axes,  draw 
the  lines  parallel  to  the  axis  of  x.  Twist  this  rectangle  as 
shown  in  the  figure,  leaving  the  line  along  the  axis  of  x  fixed 
and  sliding  each  parallel  line  along  itself  into  a  new  position. 
Thinking  of  the  lines  as  representing  the  edges  of  a  pack  of 
cards  or  of  a  block  of  paper  is  an  aid  in  visualizing  the  motion. 


352 


ANALYTIC   GEOMETRY 


COrW 


If  the  line  one  unit  above  the  a>axis   slides   to   the   right 
through  the  distance  k,  then  the  line  y  units  above  the  o>axis 

will  evidently  slide  to  the 
right  through  the  distance  ky, 
while  a  line  which  is,  say,  2 
units  below  the  ic-axis  will 
slide  a  distance  2  ft  to  the 
left.  In  other  words,  the 
algebraic  distance  through 
which  each  line  slides  is 
equal  to  k  times  the  algebraic 
distance  of  the  line  from  the 


=1=F 


FIG.  14 


'  ovaxis. 

This  is  true,  also,  of  the 
motion  of  each  point  of  the 
rectangle,  since  the  lines  slide  as  units.  If,  then,  the  whole 
plane  is  twisted  according  to  this  law,  an  arbitrary  point  (x,  y) 
will  be  carried  along  a  parallel  to  the  axis  of  x  through  the 
algebraic  distance  ky.  Hence 

tx'  =  x+ky, 

are  the  equations  of  the  transformation. 

Thus  far  we  have  i  assumed  that  k  is  positive.  It  may 
equally  well  be  negative.  Then  points  above  the  axis  of  x  are 
shifted  to  the  left,  and  points  below  it  to  the  right. 

If  the  sliding  were  along  parallels  to  the  axis  of  y,  the  trans- 
formation would  be 

(2)  lir' 


where  Z  is  any  constant,  not  zero. 

These  transformations  are  known  as  simple  shears,  and  the 
motions  which  they  generate  are  called  shearing  motions. 


Example  1.     Subject  the  curve 


(3) 


TRANSFORMATIONS  OF  THE  PLANE 


353 


to  the  shear 

Here 

(4) 

and  (3)  becomes 


x'  =  x, 


x  = 


=  2x'+y', 


2x' 


or 

(5)  y'  =  x>\ 

Conversely,  the  curve  (5)  is  carried  by  the  shear  (4)  into  the 
curve  (3).  The  shear  (4)  adds  to  the  ordinate  of  a  point 
(x1,  y'~)  the  amount  2x'  equal  to  the  cor- 
responding ordinate  of  the  line  y'  =  2  x'. 
Consequently,  the  ordinates  of  the  curve 
(3)  can  be  obtained  by  adding  to  the  ordi- 
nates of  the  line  y'  =  2x'  the  corresponding 
ordinates  of  the  curve  (5),  whose  graph  is 
known.  Thus  the  curve  (3)  can  be  easily 
plotted.  It  is  tangent  to  the  line  y  =  2  x 
at  the  origin. 


Example  2. 


(2)  l/ 


FIG.  15 


Construct  the  curve 

y  =  4  x3  —  x. 

This  is  done  by  plotting  the  line  y  =  —  x 
and  the  curve  y  =  4  a^,  and  then  adding  their  ordinates  al- 
gebraically for  a  new  ordinate  —  that  of  the  required  curve. 
The  process  is  equivalent  to  subjecting  the  curve  y'  =  4a;'3  to 
the  shear 

x  =  x',        y  =  —  x'  +  y'. 

Properties  of  Simple  Shears.  Since 
the  transformations  (1)  and  (2)  are 
special  affine  transformations  (cf.  §  6), 
it  follows  that  simple  shears  carry 
straight  lines  into  straight  lines,  parallel 
lines  into  parallel  lines,  and  tangent 
curves  into  tangent  curves.  They  do 
not  in  general  preserve  angles. 


d)j/=4a; 


FIG.  16 


354 


ANALYTIC   GEOMETRY 


FIG.  17 


Simple  shears  do,  however,  preserve 
areas.  For,  first,  this  is  true  for  any 
rectangle  whose  base  is  parallel  to  the 
direction  of  shearing,  since  such  a  rec- 
tangle is  carried  into  a  parallelogram 
with  the  same  lengths  of  base  and  alti- 
tude ;  cf.  Fig.  14.  Secondly,  the  area  A 
of  any  other  figure  can  be  considered 
as  the  limit  of  the  sum  of  the  areas  of 
rectangles  of  the  type  just  described, 
which  are  inscribed  in  the  figure  as 
shown.  But  this  sum  is  equal  always 
to  the  sum  of  the  areas  of  the  corre- 
sponding parallelograms,  whose  limit  is 
the  area  A'  of  the  transformed  figure. 


Consequently,  A  =  A',  q.  e.  d. 

EXERCISES 

Construct  the  following  curves. 

1.   y  =  2x*  +  %x.  4.  x  =  y*  +  3y. 

5.  3x  =  y3  —  6y. 

6.  2  x  =  —  4#3  —  3y. 


2.  y  =  2x3-%x 

3.  y  =  —  y?  +  x. 


The  same  for  the  following  curves,  making  use  of  a  transla- 
tion as  well  as  a  shear. 

+  x-2.  9.   x  =  2y*  —  3y  +  l. 

2x*  —  x  +  3.  10.   3x  =  —  6yz  —  2y  —  7. 

11.  Construct  the  curve 

y  =  2x3-6x2  +  7x—  1, 

beginning  by  putting  the  equation  into  the  form  : 
y  —  b  =  2(x  —  a)3  +  k(x  —  a). 

The  same  for  the  curves  : 

12.  y  =  i? 


TRANSFORMATIONS   OF   THE   PLANE 


355 


Factor  the  two  following  shears  by  the  method  of  §§  6,  7. 

% 

14.   x  =  x  -f-  -?~v  3y«  y  ==  v. 

9.   Second  Method  of  Factorization.    Homogeneous  Strains. 
THEOREM.     The  homogeneous  affine  transformation 


(1) 


f  x'  =  a  x  +  b  y, 

,        ,     ;   . ,  '         A  =  ab'  -  a'b  =£  0, 
\  y'  ==  a'x  +  b'y, 


can  be  factored  into  one-dimensional  strains  and  simple  shears, 
with  the  addition,  in  certain  cases,  of  a  reflection  in  one  or  both 
axes. 

Case  1 :  a  and  b'  not  both  0.  In  proving  the  theorem  we 
begin  with  the  case  in  which  a  and  b'  are  not  both  zero,  and 
assume  first  that  a  =f=  0.  A  simple  shear  which  suggests  itself 
as  a  probable  component  of  (1)  is 


(2) 


x1  =  x+-y, 
a 

2/i  —  y- 


Eliminating  x,  y  from  (1)  and  (2),  we  obtain 


(3) 


x'  =  axt, 

y  =  aX  +  - 

a 


This  transformation  suggests  as  a  factor  the  second  shear : 

*'  =  «2, 

'  =  kxi  +  y«, 

where  the  value  of  k  is  to  be  determined.     Elimination  of 
x',  y'  from  (3)  and  (4)  gives 


(4) 


(5) 


An  obvious  choice  for  k  is  that  which  makes  a'—  ka=  0 ;  then 
k  =  a' /a. 


356 


ANALYTIC   GEOMETRY 


We  have  now  factored  (1)  into  the  transformations  (2),  (5), 
(4),  where  k  =  a' /a,  namely  into : 

(6) 


A 

2/2  =- 
a 


The  first  and  last  of  these  transformations  are  simple  shears. 
The  second  can  be  factored  into  two  one-dimensional  strains, 
or,  in  case  a  or  A/a  or  both  are  negative,  into  these  and  a  re- 
flection in  one  or  the  other  or  both  axes.  Thus  the  theorem 
is  proved  in  this  case. 

The  proof  is  similar  in  the  case  that  b'  =£  0.  The  factors  of 
(1)  are 


(7) 


y'  =  2/2, 


2/2  = 


(8) 


Case  2  :  a  =  b'  =  0.     Here  (1)  becomes 

'  =  by, 


( y'  =  a'x. 


A  =  -  a'b  =£  0. 


This  transformation  can  be  factored  into  a  rotation  about  the 

origin  through  90° : 

(9)  «i=  — y,  2/i  =  a, 

and  the  transformation 

x'  =  —  bxl}  y'  =  ayi. 

It  can  be  shown  that  the  rotation  (9)  is  the  product  of  three 
simple  shears,  namely, 


and  this  completes  the  proof  of  the  theorem. 

Homogeneous  Strains.     The  extension  to  space  of  the  trans- 
formations (1)  is  given  by  the  formulas 
x'  =  a  x  +  b  y  +  c  z, 
y'  =  a'  x  -f  b1  y  +  c'  z, 
z'  =  a"x  +  b"y  +  c"z. 


TRANSFORMATIONS  OF  THE  PLANE  357 

The  three-dimensional  case  admits  a  treatment  similar 
to  the  foregoing,  and  the  results  are  like  those  obtained 
above. 

Transformations  of  the  form  (1)  or  (9)  are  known  in 
physics  as  homogeneous  strains.  They  are  of  particular 
importance  in  the  theory  of  elasticity.  For,  it  can  be  shown 
that,  if  an  elastic  body,  such  as  a  solid  piece  of  rubber 
or  of  steel,  is  slightly  deformed  from  its  normal  shape,  the 
displacement  of  its  points  can  be  represented  to  a  high 
degree  of  approximation  by  a  transformation  of  the  form  (1) 
or  (9). 

It  is  a  fact,  which  we  shall  not  attempt  to  prove,  that  a 
transformation  (1)  representing  in  the  above  sense  a  slight  de- 
formation cannot  have  a  reflection  in  an  axis  as  one  of  its  com- 
ponent transformations. 

One-dimensional  strains  —  simple  elongations  and  compres- 
sions —  and  simple  shears  are  often  given  the  single  name, 
simple  strains.  Adopting  this  terminology,  we  can  say  :  Every 
homogeneous  strain  representing  in  the  above  sense  a  slight  deforma- 
tion can  be  generated  by  a  succession  of  simple  strains  without 
rejections. 

EXERCISES 

Factor  the  following  homogeneous  strains  by  the  method  of 
this  paragraph. 

1.    The  strain  of  Ex.  1,  §  7.   2.    The  strain  of  Ex.  3,  §  7. 
3.   x'  =  6y,  y'  =  —  2x  +  y.       4.   x'=  3x  —  5y,  y1  =  4a;  +  3y. 


The  following  homogeneous  strains  represent  slight  dis- 
placements. Factor  them  and  note  that  a  reflection  in  an  axis 
never  appears  as  a  component  transformation. 

5.  aj'  =  1.01a?  +  .02y,  yr  =  .03z  +  .98  y. 

6.  aj'  =  .9a?-.ly,  y'=.2x  +  l.ly. 

7.  x'=(l  +  a)x  +  py,  y1  =  y  x  +  (1  +  %, 
where  a,  (3,  y,  S  are  small  quantities,  such  as  .01  or  —  .08. 


358  ANALYTIC   GEOMETRY 

Factor  the  following  transformations. 

8.  The  transformation  of  Ex.  4,  §  7. 

9.  x'  =  -2x  +  y  +  3,  y'  =  3x-2y  +  l. 

EXERCISES  ON   CHAPTER    XV 

1.  Prove  that  the  product  of  any  two  affine  transformations 
is  an  affine  transformation. 

Definition.  The  general  affine  transformation  (1),  §  6,  is 
called  non-singular,  if 

A  =  a&' -  a'&  =£  0 ; 

if  A  =  0,  it  is  called  singular.  The  expression  A  is  known  as 
the  determinant  (cf.  Ch.  XVI)  of  the  transformation. 

2.  The  inverse  of  any  non-singular   affine   transformation 
is  non-singular.     This  was  proved  incidentally  on   p.  344  in 
the  case  of  a  homogeneous  transformation.     Prove  it  in  the 
general  case. 

3.  Show  that  the  product  of  two  non-singular  affine  trans- 
formations is  non-singular. 

Suggestion.  Prove  that  the  determinant  of  the  product 
transformation  is  the  product  of  the  determinants  of  the  given 
transformations. 

4.  The  transformation, 

x'  =  2x-y,        y'  =  ±x-2y, 

is  singular.  Verify  this  and  show  that  the  transformation 
carries  all  the  points  of  the  plane  into  points  of  the  line 
2x'  —  y'  =  0.  Has  the  transformation  any  inverse  ? 

5.  The  product  of  the  general   rotation  about  the  origin, 
followed  by  the  general  translation,  is  a  transformation  known 
as  the  general  rigid  motion.     Find  its  equations. 

-4ns.   x'  =  x  cos  6  —  y  sin  6  +  a  ;  y'  =  x  sin  6  +  y  cos  6  +  b. 

6.  The  product  of  the  transformation  of  similitude  of  §  3 
and   the   general  rigid   motion  of   the   preceding   exercise  is 
known  as  the  general  transformation  of  similitude.     Find   its 


TRANSFORMATIONS  OF  THE   PLANE  359 

equations,  and  show  that  it  is  identical  with  the  general  isogonal 
transformation  for  which  b'  =  a,  a'  —  —  b  (§  6,  Th.  2). 

7.  A  non-singular  affine  transformation  carries    the   four 
collinear  points  Plt  PZ)  Qi,  Q2,  into  the  four  collinear  points 
P'i,  ?'*  Q'i>  Q'*>     Prove  that,  if  Qj,  Q2  divide  Plt  Pz  harmon- 
ically, Q'i,  Q'2  will  divide  P'iP'i  harmonically. 

Suggestion.  Prove  the  theorem  first  for  the  transformations 
considered  in  §§  1-5. 

8.  Find  the  equations  of  a  rotation  of  the  plane  about  the 
point  (x0,  y0)  through  the  angle  6. 

9.  The  plane  is  stretched  uniformly  in  all  directions  away 
from  the  point  (XQ,  y0).     Find  the  equations  representing  the 
transformation. 

10.  Deduce  the  equations  of  the  reflection  in  the  line 

Ax  +  By+  C=0. 

11.  Deduce   the   formulas   representing  a   one-dimensional 
strain  away  from  the  line  of  Ex.  10. 

12.  Find  the  equations  of   the  simple  shear  which  leaves 
each  point  of  the  line  of  Ex.  10  fixed. 

13*.  Let  the  simple  shear  (1),  §  8,  be  factored  into  the  three 
transformations  of  §  6,  namely  (10),  (16),  and  the  inverse  of 
(11).  Prove  that  sin  20'  =  sin  26,  but  that  the  only  allowable 
solutions  of  this  equation  are  0'  =  90°— 0  and  6'=  270°  -  0, 
together  with  those  equivalent  to  them.  Show  that,  if  the 
first  of  these  solutions  is  chosen,  a  =  tan  0,  8  =  cot  0,  whereas, 
if  the  second  is  taken,  a  =  —  tan  0,8  =  —  cot  0.  Prove  that,  in 
either  case,  2  cot  2  0  =  Jc. 

Suggestion.  Form  the  product  of  the  three  transformations 
and  demand  that  it  be  identical  with  the  transformation  (1), 
§8. 


CHAPTER   XVI 
DETERMINANTS   AND  THEIR   APPLICATIONS 

I.    DETERMINANTS 

1.   Simultaneous    Linear   Equations.     The   solution  of  the 
simultaneous  equations, 


y  =  kz, 
is 

/o\  „  _  kjbz  —  K20i  _ 

\r)  --  1  -  T"'  "  ~ 


provided  a^  —  «2&i  =£  0.* 

If  we  have  three  simultaneous  linear  equations  in  three 
unknowns, 

CL&  +  biy  +  c&  =  kit 
(3)  a&  +  b%y  +  c&  =  kz, 

ayx  +  b3y  +  c3z  =  k3, 

and  first  eliminate  0,  obtaining  two  equations  in  x  and  y,  and 
then  from  these  equations  eliminate  y,  we  find,  as  the  value  of  x, 

u\       x  _  kjbzCa  +  kjfyA  + 

«1&2C3  +  «2^3C1  + 

Similarly,  we  can  find  the  values  of  y  and  z.     These  will  also  be 
in  the  form  of  quotients,  with  the  same  denominator  as  in  (4), 

*If  a^bz  —  «2&i  =  0  (but  ai  and  61,  and  a2  and  62,  are  not  both  zero), 
the  two  straight  lines  represented  by  equations  (1)  are  either  parallel 
or  coincident  (Ch.  2,  §  10,  Ths.  3,  5)  ;  in  the  former  case  the  equations 
have  no  solution,  in  the  latter,  infinitely  many  solutions.  Both  cases  are 
exceptional  to  the  general  case,  ai&2  —  «2&i  =£  0,  in  which  the  solution 
(2)  is  unique. 

360 


DETERMINANTS  361 

and  the  solution  is  valid  subject  to  the  condition  that  this 
denominator  is  not  zero. 

2.  Two-  and  Three-Rowed  Determinants.  The  expressions 
in  the  numerators  and  denominators  of  the  quotients  in  (2) 
and  (4)  are  of  so  great  importance  that  they  are  given  a  name. 
They  are  called  determinants,  —  those  in  (2),  determinants  of 
the  second  order,  and  those  in  (4),  determinants  of  the  third 
order.  A  determinant,  then,  is  a  polynomial  of  the  above  type. 

The  determinant  of  the  second  order, 


can  easily  be  remembered  by  means  of  the  diagram 

+ 
(1) 


in  which  the  lines  and  the  signs  show  how  the  terms  of  the 
determinant  are  to  be  obtained. 
The  diagram 

^:x^ 


(2) 


fulfills  the  same  purpose  for  the  determinant  of  the  third  order, 


The  four  quantities  al}  ct2,  61,  &2,  arranged  in  a  square  as  in 
(1),  form  what  is  known  as  a  square  array  of  the  second  order. 
Similarly,  the  system  of  nine  quantities,  which  forms  the  basis 
of  the  diagram  (2),  is  known  as  a  square  array  of  the  third  order. 
The  square  array  is  not  itself  the  determinant.  It  is  merely 
a  convenient  arrangement  of  the  given  four,  or  nine,  quantities, 
from  which  the  value  of  the  determinant  can  be  written  down. 
However,  it  is  common  practice  to  use,  as  a  symbol  or  nota- 


362 


ANALYTIC   GEOMETRY 


tion  for  the  determinant,  the  square  array  inclosed  between 
vertical  bars,  and  to  write,  accordingly : 


a2 


(4) 


03 


These  symbols  for  the  determinants  are'  sometimes  abbreviated 
still  further.  Instead  of  the  first,  we  often  find  |  ai  62 1  or 
merely  |  a  b  |,  and  for  the  second,  |  Oj  62  cs  |  °r  |  o  6  c  (.* 

The  solution  of  the  equations  (1),  §  1,  we  can  now  write 
in  the  form 


x  = 


"•j 


or  more  compactly, 
(5) 


y  = 


The  solution  of  the  equations  (3),  §  1,  becomes 
k  b  c\  \a  k  c\  \a  b  k 


(6)   x  = 


y  = 


b  c\' 


z  = 


I      ,      , 
a  b  c 


The  value  of  a;  is  as  given  by  (4),  §  1.  The  determinant 
|  o  6  c  |  in  the  denominator  is  evidently  the  determinant  of 
the  coefficients  of  x,  y,  z  in  the  given  equations,  and  the 
determinant  \k  b  c  |  in  the  numerator  is  obtained  from  |  a  6  c  | 
by  replacing,  respectively,  a^  a2,  a3  —  the  coefficients  of  x  — 
by  klt  k2,  k3  —  the  constant  terms.  Similarly,  the  numerator 
|  a  k  c  |  of  the  value  of  y  is  obtained  from  |  a  b  c  \  by  replacing 

*  The  vertical  bars  must  not  be  confused  with  the  absolute  value  signs. 
They  have  nothing  to  do  with  these.  The  context  will  always  show 
which  meaning  is  intended. 


DETERMINANTS 


363 


the   6's  —  the   coefficients   of  y  —  by   the   k's ;    and   likewise 
for  z* 

The  four,  or  nine,  quantities  from  which  the  determinant  is 
formed  are  known  as  the  elements  of  the  determinant.  The 
rows  and  columns  in  which  they  are  arranged  are  called  the 
rows  and  columns  of  the  determinant.  The  diagonal  contain- 
ing the  elements  av,  b2  (c3)  is  the  principal  diagonal ;  the  other, 
the  secondary  diagonal.  The  determinants  are  often  called  two- 
and  three-rowed  determinants,  instead  of  determinants  of  the 
second  and  third  orders. 


EXERCISES 

Evaluate  the  following  determinants. 


1. 


5. 


2  3 

3  5 

352 
213 
437 


2. 


1    3 

0     -4 

2A       B 

3. 

4. 

-2     7 

2         7 

B     2C 

24         3 

2 

-13 

6. 

-31         2 

7. 

-1 

5    0 

15-6 

0 

3    2 

Solve  the  following  simultaneous  equations  by  means  of 
determinants.  Check  your  answers. 

Remark.  The  constant  terms  in  (1)  and  (3),  §  1,  are  on  the 
right-hand  sides  of  the  equations.  The  formulas  (5)  and  (6), 
for  the  solution  of  the  equations,  are  subject,  then,  to  the 
arrangement  of  the  equations  in  this  form. 

2  x  —  y  =  3, 

3  x  +  4  y  =  5. 


8. 


9. 


10. 


11. 


12.   5x—  Sy  +  z  —  9  = 


2z  —  7  =  0, 


*  Formulas  (6)  are  proved  later,  in  §  8. 


364 


ANALYTIC   GEOMETRY 


3.   Determinants  of  the  Fourth  and  Higher  Orders.     Given 
sixteen  quantities  arranged  in  a  square  array  : 

tti     61     GI    d\ 

xv  02       62       C2       ^2 


at    64     c4     dt 
What  shall  we  mean  by  the  determinant  symbolized  or  de- 


noted by 


&! 


0*2 


a3 


c3 


a4    &4     c4 

or,  more  simply,  by  |  a^  b2  <%  d±  \     or     |  a  6  c  d  \? 

If  we  were  to  proceed  as  before,  we  should  write  down  four 
simultaneous  linear  equations  in  four  unknowns,  with  the 
elements  of  (1)  as  the  coefficients  of  the  unknowns  and  fc1}  k2, 
A;3,  k4  as  the  constant  terms,  and  then  solve  the  equations. 
The  value  of  each  unknown  would4be  a  quotient,  and  all  four 
quotients  would  have  the  same  denominator,  which  we  should 
then  define  as  the  determinant  |  a  b  c  d  |.  As  a  matter  of  fact, 
this  denominator  and  each  of  the  numerators  contains  24  terms. 
The  prospect  of  solving  the  equations  is,  then,  forbidding. 

Why  not  form  the  products  suggested  by  a  diagram"  based  on 
(1),  similar  to  the  diagram  for  the  three-rowed  determinant,  pre- 
fix the  proper  signs,  and  call  the  result  the  determinant  ?  Un- 
fortunately this  method  yields  but  8  terms,  whereas  according 
to  our  prediction  the  determinant,  properly  defined,  contains  24. 

We  adopt  here  a  new  method  of  attack.  Let  us  inspect 
more  closely  the  relationship  between  the  square  arrays  of 
orders  two  and  three  and  the  corresponding  determinants. 
Consider  a  specimen  term  of  the  determinant  (4),  §  2.  It  con- 
tains just  one  a,  just  one  6,  and  just  one  c  ;  furthermore,  each  of 
the  subscripts  1,  2,  3  appears  just  once.  In  other  words,  the 
term  is  the  product  of  three  elements,  one  from  each  row  and  one 


DETERMINANTS  365 

from  each  column  of  the  square  array.  Moreover,  every  product 
of  this  type  is  present  as  some  term  in  the  determinant,  as  can 
be  shown  by  writing  down  all  such  products  and  comparing 
them  with  the  terms  of  the  determinant. 

By  analogy,  then,  to  form  the  determinant  \a  b  c  d\,  we 
should  write  down  all  the  products  of  elements  of  (1),  each  of 
which  contains  just  one  factor  from  each  row  and  just  one 
factor  from  each  column  of  (1),  that  is,  all  the  products  of  the 
form  Gi&yC/Z,,  where  i,  j,  Jc,  I  are  the  numbers  1,  2,  3,  4  in  all 
possible  orders.  There  are  24  such  products.  For,  we  can 
choose  the  first  factor,  say  from  the  column  of  a's,  in  four  ways 
—  from  any  one  of  the  four  rows ;  and  then  the  second  factor, 
say  from  the  column  of  6's,  in  three  ways  —  from  any  one  of  the 
three  remaining  rows  ;  and  the  third  factor,  in  two  ways  ;  the 
fourth  is  then  uniquely  determined.  The  number  of  possible 
products  is,  therefore,  4  •  3  •  2  •  1  =  4  !  =  24. 

It  remains  to  determine  the  signs  to  be  given  to  the  24 
products.  Toward  this  end,  let  us  write  down  the  subscripts  of 
the  terms  of  (4),  §  2,  in  the  order  in  which  they  occur,  ivhen 
the  letters  a,  b,  c  are  in  their  natural  order.  For  the  terms  with 
plus  signs  we  have 

123,        231,        312, 
and  for  the  terms  with  minus  signs, 

321,         213,        132. 

The  first  set  1  2  3  is  normal.  In  the  second  set,  2  3  1,  2  and 
3  each  precede  1,  and  we  say  that  there  are  two  inversions  from 
the  normal  order.  In  3  1  2,  3  precedes  1  and  2,  —  again  two 
inversions.  In  the  three  sets  for  the  negative  terms  the  num- 
ber of  inversions  is  respectively  three,  one,  and  one. 

It  appears,  then,  that  the  number  of  inversions  in  the  set  of  sub- 
scripts for  a  term  with  a  plus  sign  is  even  (or  zero),  whereas  for  a 
term  with  a  minus  sign,  this  number  is  always  odd. 

Proceeding  according  to  this  rule,  we  should  give  to  each  of 
the  24  products,  a.&^d,,  formed  from  (1)  a  plus  sign  or  a  minus 


366  ANALYTIC   GEOMETRY 

sign,  according  as  i  j  k  I  presents  an  even  or  an  odd  number  of 
inversions  from  the  normal  order  1234.  Thus  the  product 
&2d1a3c4  would  be  taken  as  plus,  since,  when  the  factors  are 
arranged  in  the  order  of  the  letters,  viz.  —  a362c4c?i,  the  num- 
ber of  inversions  in  the  subscripts  3  2  4  1  is  even,  namely  4. 
The  product  afi&di  would  be  taken  as  minus,  since  the  num- 
ber of  inversions  in  4  2  3  1  is  odd,  namely  5. 

We  can  now  give  a  complete  definition  of  the  determinant  of 
the  fourth  order. 

DEFINITION.  Form  all  the  products  of  elements  of  (1)  which 
contain  just  one  factor  from  each  row  and  one  factor  from  each 
column  of  (1)  ;  to  each  product  af&yCfcd,  prefix  a  plus  sign  or  a 
minus  sign,  according  as  the  number  of  inversions  of  i  j  k  I  from 
the  normal  order  1  2  3  4  is  even  or  odd.  The  sum  of  the 
products,  thus  signed,  is  the  determinant. 

Determinants  of  the  fifth,  sixth,  and  higher  orders  are 
similarly  defined.  Let  the  student  think  through  the  definition 
for  a  five-rowed  determinant,  and  let  him  show,  also,  that  in  the 
case  of  two-and  three-rowed  determinants  the  definition  yields 
precisely  the  expressions  which  were  defined  as  these  determi- 
nants in  §  2. 

The  signed  products  which  make  up  a  determinant  are  known 
as  the  terms  of  the  determinant.  Thus,  +  o^c^i  and 
i  are  terms  of  |  a  b  c  d  \. 


EXERCISES 

1.  What  is  the  number  of  inversions  of  each  of  the  following 
orders,  from  the  normal  order  ? 

(a)  3  1  4  2  ;     (c)  2  5  3  1  4  ;      (<?)    3  1  6  4  5  2  ; 
(6)  2  4  3  1  ;     (d)  4  3  5  2  1  ;     (/)  6  5  4  3  2  1. 

2.  Write  out  all  the  terms  of  |  a  6  c  d  \.   -To  how  many  pro- 
ducts have  you  prefixed  plus  signs  ?   To  how  many,  minus  signs  ? 

3.  How  many  terms  has  a  determinant  of  the  fifth  order  ? 
Prove  your  answer. 


DETERMINANTS 
4.    The  same  for  a  determinant  of  the  nth  order. 


367 


5.  Show  that  the  sign  to  be  prefixed  to  the  product  of  the 
elements  of  the  principal  diagonal  is  always  the  plus  sign, 
no  matter  what  the  order  of  the  determinant. 

4.  Evaluation  of  a  Determinant  by  Minors.  Fix  the  atten- 
tion on  a  particular  element  of  a  determinant  A.  Cross  out 
the  row  and  column  in  which  this  element  stands.  There  will 
remain  a  determinant  of  order  one  less  than  that  of  A.  This 
determinant  is  known  as  the  minor  of  the  element  chosen. 

For  example,  the  minor  of  a2  in  the  determinant 


(1) 

is  the  determinant 


Consider  the  product  a2A.,.  The  terms  in  this  product  are 
and  —  a2&3ci,  and  by  (4),  §  2,  these  are  terms  of  (1)  ex- 
cept for  sign;  moreover,  they  are,  except  for  sign,  all  the 
terms  of  (1)  which  contain  a2. 

Again,  the  terms  of  b.2B<>,  where  B%  is  the  minor  of  62  in  (1), 
are  a-f>^  and  —  ajb^.  These  are  precisely  terms  of  (1)  and, 
in  fact,  all  the  terms  of  (1)  which  contain  62. 

In  general,  let  tn  be  an  element  of  a  determinant  A  and  let 
M  be  its  minor.  Then  the  terms  of  the  product  mM  are  terms 
of  A,  except  perhaps  for  sign ;  furthermore,  they  are,  except 
perhaps  for  sign,  all  the  terms  of  A  which  contain  m. 

For,  if  we  take  the  factor  m  from  a  term  of  A  which  con- 
tains m,  the  product  which  remains  contains  just  one  element 
from  each  row  and  column  other  than  the  row  and  column  in 
which  m  stands,  and  is,  therefore,  a  product  occurring  in  the 
determinant  M.  And,  if  we  take  the  factor  m.  from  all  the 
terms  of  A  containing  m,  the  products  which  remain  are  all 


368  ANALYTIC   GEOMETRY 

the  products  of  the  type  described  and  hence  are  all  the 
products  occurring  in  M,  q.  e.  d. 

It  will  be  shown  later  (§  7)  that  the  terms  of  m M  as  they 
stand,  or  the  terms  of  mM  with  all  their  signs  changed,  are 
precisely  terms  of  A,  according  as  the  sum  of  the  number  of 
the  row  and  the  number  of  the  column  in  which  m  stands  is 
even  or  odd.  Assuming  this,  we  can  now  state  the  theorem  : 

THEOREM  1.  If  m  is  the  element  in  the  i-th  row  and  j-th  column 
of  A,  and  M  is  its  minor,  +  mM  or  —  mM,  according  as  i  +  j 
is  even  or  odd,  consists  of  all  the  terms  of  A  which  contain  m. 

Thus,  in  the  case  of  the  element  a2  of  (1),  i  =  2,j  =  1,  and 
i  -}-  j  =  3 ;  accordingly,  —  a^Az  gives  all  the  terms  of  (1)  con- 
taining 02.  For  62>  *  =  2,  j  =  2,  and  i  +  j  =  4,  and  so  +  b2B2 
consists  of  all  the  terms  of  (1)  containing  &2.*  Similarly,  if 
(72  is  the  minor  of  <%,  —  <%€%  consists  of  all  the  terms  of  (1) 
containing  c%. 

The  sum 

(2)  -  a2^l2  +  b2B2  -  C2<72 

is  precisely  the  value  of  the  determinant  (1).  For,  it  consists 
of  all  the  terms  of  (1)  containing  a^  or  62  or  c-j,  i.e.  containing 
an  element  of  the  second  row,  and  every  term  of  (1)  contains 
such  an  element.  The  student  should  also  verify  the  state- 
ment by  comparing  the  terms  of  (2),  when  expanded,  with 
those  of  (1). 

In  (2)  we  have  the  sum  of  the  products  of  the  elements  of 
the  second  row  by  their  minors,  each  product  having  the 
proper  sign  according  to  Theorem  1.  We  say  that  (2)  is  the 
evaluation  or  expansion  of  the  determinant  (1)  by  the  minors  of 
the  second  row. 

Similarly,  the  sum, 

(3)  c1<71-c2C2  +  c3C3, 

of  the  products  of  the  elements  of  the  third  column  of  (1)  by 
their  minors,  where  the  signs  have  been  determined  by  Theorem 

*  Compare  these  results  with  those  obtained  directly  at  the  beginning 
of  the  section. 


DETERMINANTS 


369 


1,  is  precisely  the  determinant  (1).  We  speak  of  (3)  as  the 
evaluation  of  (1)  by  the  minors  of  the  third  column. 

The  reasoning  here  is  perfectly  general,  applying  to  a  deter- 
minant of  any  order  and  to  any  row  or  column  of  the  deter- 
minant. The  result  we  summarize  as  follows  : 

EVALUATION     OF     A    DETERMINANT     BY    THE     MlNORS     OF     A 

Row  OR  A  COLUMN.  Single  out  a  row  or  a  column  of  a 
determinant.  Multiply  each  element  of  it  by  the  minor  of 
the  element  and  prefix  to  the  product  the  proper  sign,  as 
determined  by  Tfieorem  1.  The  sum  of  the  signed  products  is 
the  determinant. 

We  now  have  a  feasible  means  of  finding  the  values  of  de- 
terminants of  the  fourth  and  higher  orders.  For  example,  the 
determinant 


2 
4 

-3 
-1 


5 

-9 
6 
4 


-2 
3 

-4 
-3 


8 

-7 
4 
5 


evaluated  by  the  minors  of  the  first  row,  is  equal  to 


-9  3  -7 
6-4  4 
4-3  5 


-5 


3   -7 


-3  -4 
-1  -3 


+  (-2) 


4  -9  -7 


-3 

-1 


-8 


4-9      3 

-3 

6  -4 

-1 

4  -3 

When  the  value  of  each  of  the  three-rowed  determinants  is 
computed,  by  the  above  method  or  by  that  of  the  diagram  of 
§  2,  this  becomes 

2(44)-5(-34)-2(-l)-8(19), 

which  yields  finally,  as  the  value  of  the  given  determinant, 
108. 

EXERCISES 

1.  Given  the  determinant  \a  b  c\  and  the  three  quantities 
ki;  k^,  ks.  Prove  that 

kiAl  —  k2Az  +  k3A~  =\k  b  c\. 


370 


ANALYTIC   GEOMETRY 


2.  (Generalization  of  Ex.  1.)  Given  a  determinant  A  of 
the  nth  order  and  the  n  quantities  ki}  k2,  •  •  •,  kn.  Single  out  a 
column  (or  row)  of  A,  form  the  minors  of  its  elements  and 
prefix  to  each  the  sign  prescribed  by  Th.  1.  Multiply  each 
signed  minor  by  the  corresponding  k  and  take  the  sum  of  these 
products.  Prove  that  this  sum  is  equal  to  the  determinant 
obtained  from  A  by  replacing  the  column  (or  row)  in  question 
by  &!,  k2,  •  ••,  kn. 

By  the  method  of  this  section,  evaluate  the  following  de- 
terminants 
3. 


5. 


That 

of  Ex.  5, 

§2. 

4. 

That 

of  Ex. 

7,  § 

2. 

2 

3         1 

5 

3 

-2 

5 

1 

5 

2     -2 

1 

4 

-3 

7 

-2 

3 

4         6 

2 

• 

6. 

-6 

2 

—  3 

0 

-1 

5         2 

3 

5 

3 

-2 

2 

1         0 

-2 

0 

0 

2     -1 

4 

3 

2 

-5         2 

0 

2     - 

1 

. 

0         3 

-1 

4 

5 

2         0 

3 

1        4 

7. 


5.   Simplified  Evaluation  by  Minors.     Given  the  determinant, 

3  201 
2-134 

4  502' 
2         603 

Three  of  the  four  elements  in  the  third  column  are  zero. 
Accordingly,  if  we  expand  the  determinant  by  the  minors  of 
the  third  column,  three  of  the  four  resulting  products  have 
zero  factors  and  drop  out,  so  that  there  is  left,  merely, 

321 
-345     2  . 
263 

Hence  —  3  (7)=  —  21  is  the  value  of  the  determinant. 


DETERMINANTS 


371 


It  is  clear  from  this  example  that  a  determinant  which  has 
the  property  that  all  but  one  of  the  elements  in  some  row  or 
in  some  column  are  zero  is  very  simply  evaluated.  Conse- 
quently, if  a  determinant  which  has  not  this  property  can  be 
transformed  into  an  equal  determinant  which  has  the  prop- 
erty, a  simple  method  is  at  hand  for  the  evaluation  of  all 
determinants. 

The  transformation  in  question  is  always  possible.  It  is 
based  on  the  following  theorem. 

THEOREM  2.  If  the  elements  of  a  row  (or  column)  of  a  de- 
terminant are  each  multiplied  by  the  same  quantity  and  are  then 
added  to  the  corresponding  elements  of  a  second  row  (or  column), 
the  value  of  the  determinant  is  unchanged. 

Let  us  first  try  to  appreciate  the  value  of  the  theorem, 
postponing  the  proof  until  later.  Consider  the  determinant 
(4)  of  §  4,  namely, 


(1) 


2 

4 

-3 


5 

-9 
6 
4 


-2 
3 

_  4 

-3 


8 

-7 
4 
5 


By  application  of  the  theorem  we  proceed  to  transform  this 
determinant  into  an  equal  determinant  with  the  first  three  ele- 
ments of  the  first  column  all  zero. 

Rewrite  (1),  putting  in,  to  begin  with,  only  the  last  row : 


(2) 


-14-35 


Multiply  the  elements  of  the  last  row  of  (1)  by  2  and  add  the 
numbers  obtained  to  the  elements  of  the  first  row  of  (1) ;  the 
result  is  0,  13,  —  8,  18  as  a  new  first  row,  to  be  put  into  (2). 
Similarly,  multiply  the  last  row  by  4  and  add  to  the  second 
row,  thus  getting  0,  7,  —  9,  13  as  a  new  second  row,  to  be 
written  in  (2).  Finally,  the  last  row  multiplied  by  —  3  and 


372 


ANALYTIC   GEOMETRY 


added  to  the  third  row  gives  0,  —  6,  5,  —  11  as  a  new  third 
row.     Thus  (2)  has  become 

18 
13 

-11 
5 


(3) 


-9 

5 

-3 


0       13 

0         7 
0     -6 
-1         4 

a  determinant  whose  value,  by  Th.  2,  is  equal  to  that  of  (1). 
Expansion  of  (3)  by  the  minors  of  the  first  column  gives 

13-8         18 
_(_!)       7-9         13- 

-  6         5-11 

The  evaluation  of  this  three-rowed  determinant  by  means  of 
the  schematic  diagram  of  §  2  involves  the  multiplication  of  large 
numbers.     This  may  be  avoided  as  follows.     Apply  Theorem  2 
so  as  to  introduce  1  or  —  1  as  an  element ;  for  instance,  by  multi- 
plying the  last  row  through  by  2  and  adding  to  the  first  row : 
1         2-4 
7-9         13  • 

-  6         5-11 

Now  rewrite  the  determinant,  putting  in  just  the  first  column. 
Multiply  the  first  column  by  —  2  and  add  to  the  second 
column,  for  a  new  second  column ;  and  multiply  the  first 
column  by  4  and  add  to  the  third  column,  for  a  new  third 
column.  The  result  is  the  equal  determinant 

100 

7-23         41  , 

-  6         17-35 

which,  on  expansion  by  the  minors  of  the  first  row,  has  the  value 
-23         41 

17     -35  ' 

or  (-  23) (-  35)-  17  .  41  =  805  -  697  =  108.* 

Thus  the  determinant  (1)  has  the  value  108. 

*  Here,  too,  the  long  multiplications  could  be  replaced  by  simpler  ones, 
through  the  application  of  the  above  method  of  reduction. 


DETERMINANTS 


373 


EXERCISES 

Evaluate,  by  the  above  method,  the  following  determinants. 

57-3 


897 
1.     7     6    5  .  2. 

,074 
4..  That  of  Ex.  5,  §  4. 
6.    That  of  Ex.  7,  §  4. 


10 
21 
30 


12 
26 
33 


15 
30 
37 


3. 


-6 


5.    That  of  Ex.  6,  §  4. 


6.  Fundamental  Properties  of  Determinants.  The  following 
theorems  are  fundamental  in  the  transformation  and  evalua- 
tion of  determinants.  They  lead  up  to  a  simple  proof  of 
Theorem  2. 

THEOREM  3.  If  all  the  elements  of  a  row  (or  column)  are 
multiplied  by  the  same  quantity,  the  value  of  the  determinant  is 
multiplied  by  this  quantity. 

For,  each  term  of  a  determinant  A  contains  as  a  factor  just  one 
element  from  the  row  (or  column)  in  question,  and  consequently, 
when  the  elements  of  this  row  (or  column)  are  all  multiplied 
by  the  same  quantity,  m,  the  terms  of  A  will  all  be  multiplied 
by  m,  and  the  resulting  determinant  will  have  the  value  mA. 

The  theorem  is  often  of  use  in  evaluating  a  determinant. 
For  example,  if  A  is 


then 

A  =  7 


5      2 
20    42 


=  7-5 


35     14 

20    42 

> 

1      2 
4    42 

=  7-5-2 

1       1 
4     21 

=  70  •  17  =  1190. 


THEOREM  4.     If  two  rows  (or  columns)  of  a  determinant  are 
identical,  element  for  element,  the  determinant  has  the  value  zero. 
Let  us  assume  that  it  is  two  rows  which  are  identical.     The 
proof  is  similar,  if  it  is  two  columns. 

For  a  determinant   of   the   second   order,  the   theorem   is 
<  obvious : 

a    b 


a    b 


=  ab  —  ab  =  0. 


374  ANALYTIC   GEOMETRY 

Consider,  next,  a  determinant  of  the  third  order,  with  two 
rows  identical.  Expand  the  determinant  by  the  minors  of 
the  third,  or  odd,  row.  Each  of  these  minors  has  its  two  rows 
identical  and  is,  therefore,  zero,  since  the  theorem  has  been 
proved  for  two-rowed  determinants.  Consequently,  the  given 
determinant  is  zero. 

Similarly,  having  proved  the  theorem  for  three-rowed"  de- 
terminants, we  can  prove  it  for  a  four-rowed  determinant. 
For,  we  have  but  to  expand  the  four-rowed  determinant  by  the 
minors  of  a  row  which  is  not  one  of  the  two  identical  rows. 
This  expansion  will  have  the  value  zero,  since  each  of  the 
minors  in  question  is  a  three-rowed  determinant  with  two 
identical  rows. 

The  process  perpetuates  itself.  Hence  the  theorem  is  true 
for  a  determinant  of  any  order. 

The  method  of  proof  used  here  is  known  as  mathematical 
induction.  The  fact  that  the  theorem  is  true  for  a  two-rowed 
determinant  leads  up  to  its  truth  for  a  three-rowed  deter- 
minant, etc. 

COKOLLARY.  If  the  elements  of  two  roius  (or  columns)  of  a 
determinant  are  proportional,  the  determinant  has  the  value 
zero. 

For,  each  element  of  one  of  the  two  rows  (or  columns)  in 
question  is  by  hypothesis  a  multiple,  ra,  of  the  corresponding 
element  of  the  other.  Thus  m  can  be  taken  out  from  the  first 
of  the  two  rows  (or  columns)  as  a  factor  (Th.  3).  The  two 
rows  (or  columns)  are  then  identical,  and  Theorem  4  can  be 
applied. 

THEOKEM  5.  If  each  element  of  a  roiv  (or. column)  is  the  sum 
of  two  quantities,  the  determinant  can  be  written  as  the  sum  of 
two  determinants. 

Denote  the  determinant  by  A  and  the  elements  of  the  col- 
umn (or  row)  in  question  by  mt  +  m/,  m?  -\-  m2',  •••.  Denote  by 
A  the  determinant  obtained  by  replacing  all  the  m"s  in  A  by 
zeros,  and  by  A'  the  determinant  obtained  by  replacing  all  the 


DETERMINANTS 


m's  in  A  by  zeros.     We  shall  prove  that 

A  =  A  +  A'. 
For  example, 

i  +  wh'     h 
m2  -+-  m2'     62 

3  +  Wls'       63       Cs  W13       63       C3 


m2' 
m3' 


Proof.  Every  term  of  A,  since  it  contains  just  one  element 
from  the  column  (or  row)  under  discussion,  is  the  sum  of  two 
quantities,  one  containing  an  m  and  the  other  an  m'.  All  the 
quantities  containing  m's  form  the  determinant  A,  and  all  those 
containing  m"s,  the  determinant  A'.  Hence,  A=  A  +  A'. 

Or,  expand  A  by  the  elements  of  the  column  (or  row)  in 
question,  denoting  the  minors  of  these  elements  by  MI,  M2,  •«•. 
The  result  is 


(1)      A  =  ± 
=  ± 


+  m/)  MI  -  (m2  +  m2')  M2  + 
+  •••  ]  ±  [mi  MI  — 


-f-  •••  ]• 


The  values  of  A  and  A  can  be  obtained  from  (1)  by  replacing, 
first,  the  m"s,  and  then  the  m's,  by  zeros  : 


A  = 
Hence 


—  m2  M2 


»  ],          A  = 
A  =  A  +  A'. 


The  proof  of  Theorem  2,  §  5,  is  now  simple.  The  deter- 
minant A',  which  is  obtained  from  the  given  determinant  A 
by  adding  to  the  elements  ml5  wio.,  •••  of,  let  us  say,  a  column 
the  corresponding  elements  pi,  p2,  —  of  a  second  column,  each 
multiplied  by  a  quantity  fc,  contains  the  column  mi  +  Jcpi, 
m2  +  kp2,  "'.  Hence  A'  equals  the  sum  of  two  determinants, 
the  first  of  which  is  A.  The  second  has  the  two  columns 
Jcpi,  kp2,  —  and  ply  p2)  ...  and  is  therefore  zero  (Th.  4,  Cor.). 
Consequently,  A'  =  A. 

For  example,  if  A  is  the  three-rowed  determinant  (4),  §  2, 
and  to  the  elements  of  the  second  column  are  added  those  of 
the  first,  each  multiplied  by  ft,  we  have 


376 


ANALYTIC   GEOMETRY 


A'  = 


+ 


a3    63  +  ka3   c3 


a2. 


TCO.I 
A;a2 


a3 


=  A  +  0  =  A. 


EXERCISES 


1.  Prove  the  theorem  :  If  all  the  elements  of  a  row  (or  column) 
are  zero,  the  determinant  has  the  value  zero. 

2.  Given  the  determinant  |  a  b  c  | .     Using  Ex.  1,  §  4,  show 
that 

biAi  —  b2A2  -}-  63^3  =  0. 

3.  (Generalization  of  Ex.  2.)     If  to  the  minors  of  a  column 
(or  roiu)  of  ~a  determinant  are  prefixed  the  signs  prescribed  by 
Theorem  1,  and  if  each  signed  minor  is  then  multiplied  by  the 
corresponding  element  of  a  different  column  (or  row),  the  sum 
of  the  resulting  products  has  the  value  zero.     Prove  this  theorem. 
Cf.  Ex.  2,  §  4. 

Evaluate  the  following  determinants,  making  as  much  use 
as  possible  of  Ths.  2-5. 


4. 


866 
292 
661 


5. 


2 
-4 


-2 
24 


16    20       16 


6. 


4  15  -6 
6  12  9 
2  38  -3 


7.   Interchanges  of  Rows  and  of  Columns.     Given  the  first  n 
integers  in  natural  order : 

(1)  1  2  3  •  •  •  I  I  +  1  •  •  •  n ; 

in  this  order  (1)    interchange   two  successive  integers,  I  and 
l  +  l: 

(2)  123...?  +  lZ...n. 

Consider,  now,  the  n  integers  in  an  arbitrary  order, 

(3)  p  q  r  •  .  •  • t, 

and  compare  the  number  of  inversions  of  this  order  from  the 
order  (2)  with  the  number  of  its  inversions  from  the  order  (1). 


DETERMINANTS 


377 


Each  pair  of  integers  is  in  the  same  order  in  (2)  as  it 
was  in  (1),  except  the  pair  I,  I  + 1.  Consequently,  if  a  pair 
of  integers  in  (3),  not  the  pair  I,  I  -f  1,  presents  an  inversion 
from  the  order  (1),  it  also  presents  an  inversion  from  the 
order  (2),  and  vice  versa.  But  the  pair  I,  I  +  1  in  (3)  presents 
an  inversion  from  one  of  the  orders,  (1)  and  (2),  and  not  from 
the  other.  Hence,  we  conclude : 

LEMMA  1.  The  total  number  of  inversions  from  the  order  (2), 
which  (3)  presents,  differs  by  one  from  the  total  number  of  inver- 
sions from  the  order  (1),  which  it  presents. 

For  example,  23145  has  two  inversions  from  the  natural 
order  12345  and  three  from  the  order  12435. 
In  the  general  determinant  of  the  nth  order, 


A  = 


02 


Cj 


*>n        Cn         •         •         •         •         kr 

the  normal  order  for  the  subscripts  is  the  order  of  the  rows, 
namely  the  order  (1).  Accordingly,  if  in  A  two  adjacent  rows, 
the  Zth  and  (I  +  l)st,  are  interchanged,  the  normal  order  for 
the  subscripts  in  the  new  determinant,  A',  is  the  order  (2). 

The  terms  in  A  and  in  A'  are  the  same,  except  perhaps  for 
sign.  To  determine  the  sign  of  a  term  ajbqcr  •••  kt,  as  a  term 
of  A,  the  number  of  inversions  which  the  subscripts 


(3) 


p  q  r 


t 


present  from  the  order  (1)  is  counted ;  to  determine  its  sign, 
as  a  term  of  A',  the  number  of  inversions  of  (3)  from  the  order 
(2)  is  counted.  We  have  just  shown  that  the  two  results 
differ  always  by  unity.  Consequently,  the  term  in  question 


378  ANALYTIC   GEOMETRY 

has  one  sign  in  A  and  the  opposite  sign  in  A'.  Therefore, 
A'  =  —  A.  We  have  thus  proved  the  theorem : 

THEOREM  6.  If  two  adjacent  rows  of  A  are  interchanged,  the 
sign  of  A  is  changed. 

Suppose,  now,  that  we  carry  a  row  over  m  rows.  This  can 
be  effected  by  m  interchanges  of  adjacent  rows ;  for  example, 
if  the  row  is  to  be  carried  downward,  by  interchanging  it  with 
the  row  just  below  it,  then  with  the  row  just  below  its  new 
position,  etc.  Since  each  interchange  of  adjacent  rows  changes 
the  sign  of  A,  the  final  determinant  will  be  equal  to  A  or  —  A, 
according  as  m  is  even  or  odd.  This  result  we  state  in  the 
form  of  a  theorem : 

THEOREM  7.  If  a  row  of  A  is  carried  over  m  rows,  the  result 
is  A  or  —  A,  according  as  m  is  even  or  odd. 

Finally,  interchange  any  two  rows.  If  there  are  m  rows 
between  the  two,  the  interchange  can  be  effected  by  carrying 
one  of  the  rows  over  these  m  and  then  by  carrying  the  second 
one  over  this  one  and  the  m,  i.e.  over  m  + 1  rows.  Thereby 
the  determinant  experiences  m  +  m  +  l=2?n,-fl  changes  of 
sign,  i.e.  an  odd  number.  Thus  we  have  the  result: 

THEOREM  8.  If  any  two  rows  of  A  are  interchanged,  the  sign 
of  A  is  changed. 

New  Rules  for  Determining  the  Sign  of  a  Term.  We  first 
state  the  following  lemma : 

LEMMA  2.     Take  the  first  n  integers  in  an  arbitrary  order: 

(4)  p  q  r  •  •  -  i  j  •  -  •  t, 

and  in  this  order  interchange  two  ADJACENT  integers,  i  andj: 

(5)  p  q  r  •  •  •  j  i  •  •  •  t. 

Then  the  number  of  inversions  of  (5)  from  the  natural  order 
differs  from  the  number  of  inversions  of  (4)  from  the  natural 
order  by  one. 

The  proof  of  this  lemma  is  exactly  like  the  proof  of  Lemma  1. 


DETERMINANTS  379 

An  arbitrary  term  of  A,  without  its  sign,  can  be  written  in 
the  form 

(6)  vpwqxr za 

where 

(7)  v  w  x z 

are  the  letters  a  b  c A;  in  some  order  and 

(8)  p  q  r * 

are  the  subscripts  123 n  in  some  order.  Let  N  be 

the  number  of  inversions  of  the  letters  (7)  from  the  natural 
order  and  let  M  be  the  number  of  inversions  of  the  subscripts 
(8)  from  the  natural  order.  N+  M  is  the  total  number  of 
inversions  in  letters  and  subscripts. 

If  we  interchange  two  adjacent  factors  in  (6),  the  effect  is  to 
interchange  two  adjacent  letters  in  (7)  and  two  adjacent  sub- 
scripts in  (8).  Hence,  t»y  Lemma  2,  N  is  changed  by  one  * 
and  M  by  one  ;  consequently,  the  sum  N  +  M  is  changed  by 
2  or  left  unchanged.  But  any  reordering  of  the  factors  in  (6) 
can  be  effected  by  a  number  of  interchanges  of  adjacent  fac- 
tors. It  follows,  then,  that  any  reordering  of  the  factors  of 
(6)  changes  N  +  M  by  an  even  number  or  leaves  it  unchanged. 

That  is,  the  evenness  or  oddness  of  the  total  number  of  inver- 
sions in  letters  and  subscripts  in  a  term  of  A  is  independent  of 
the  order  of  the  factors  in  the  term. 

We  may,  therefore,  arrange  the  factors  with  the  letters  in 
the  natural  order  and  count  the  inversions  in  the  subscripts, 
as  in  the  definition,  §  3,  or  we  may  arrange  the  factors  with 
the  subscripts  in  the  natural  order  and  count  the  inversions 
in  the  letters,  or  we  may  leave  the  factors  unarranged  and 
count  the  inversions  in  both  letters  and  subscripts.  The  re- 
sult will  always  be  even  or  always  be  odd,  no  matter  which 
of  the  three  methods  is  used,  and  consequently  the  sign  to 
be  given  to  the  term  will  always  turn  out  to  be  the  same. 

*  Lemma  2  is  stated  in  terms  of  integers  ;  it  holds  equally  well  for 
letters. 


380  ANALYTIC   GEOMETRY 

In  the  above  methods  of  determining  the  sign  of  a  term  the 
letters  and'  subscripts  (or  the  columns  and  rows)  enter  sym- 
metrically. The  columns  and  rows  also  play  the  same  roles 
in  the  choice  of  the  factors  which  constitute  the  term.  In 
other  words,  the  formation  of  a  determinant  from  its  square 
array  bears  equally  on  the  rows  and  columns  of  the  array. 
We  have,  then,  the  following  theorem. 

THEOREM  9.  If  the  rows  and  columns  of  A  are  interchanged, 
A  is  unchanged. 

Consequently,  Theorems  6,  7,  and  8,  which  have  been  proved 
for  rows,  are  true  also  for  columns. 

Completion  of  the  Proof  of  Theorem  1.  If  m  is  the  element 
in  the  ith  row  and  jth  column  of  A,  we  have  to  show  that 
-f-  m  M  or  —  m  M  gives  terms  of  A,  according  as  i  -f  j  is  even 
or  odd  ;  or  more  briefly,  that  (—  l)*'+%Jlf  always  gives  terms 
of  A. 

If  i  =  1,  j  =  1,  i.e.  if  m  is  the  element  in  the  upper  left-hand 
corner  of  A,  the  natural  orders  of  letters  and  subscripts  in  M 

are 

b  c  •  •  •  •  Jc        and         2  3  •  •  •  •  n. 

A  term  T  of  M  will  present  the  same  number  of  inversions  in 
letters  and  subscripts  with  respect  to  these  orders  as  the  cor- 
responding'term,  aiT,  of  A  presents  with  respect  to  the  orders 

a  b  c  •  •  •  •  k        and         123 n. 

Hence  it  is  +  mM  which  gives  terms  of  A  and,  since  i  +j  =  2, 
this  is  in  accordance  with  the  theorem. 

Consider,  now,  the  general  case:  m  in  the  ith  row  and  jth 
column.  Carry  the  ith  row  over  i  —  1  rows  to  the  top  of  A 
and  then  carry  the  jth  column  over  j  —  i  columns  to  the  ex- 
treme left  of  A.  By  Th.  7,  the  resulting  determinant  is 

(9)  (-  l)<+'-'A, 

and  in  it  m  is  in  the  upper  left-hand  corner.  It  follows,  then, 
from  the  case  first  considered,  that  +  mM  gives  terms  of  (9). 


DETERMINANTS  381 

Hence  (—  l)'+'mJ/  gives  terms  of 


and  therefore  of  A,  since  2  i  +  2j  —  2  is  even,  q.  e.  d.* 

EXERCISES 

1.  Prove  Lemma  2. 

2.  Determine,  by  each  of  the  three  methods  above  described, 
the  signs  to  be  given  to  the  following  products  : 

(a)         63Cia25  (&)         CsMAj  (c) 


8.  Cramer's  Rule.  In  §  2,  we  stated  that  the  three  simul- 
taneous equations  in  three  unknowns, 

a&  +  biy  +  <?!«  =  klt 
(1)  a*x  -f  &;#  +  c-jz  =  fcj, 

030;  +  6jy  +  c32  =  k3, 
have  the  solution 

_|fc6c|  _  |  a  fc  c  |  _  I  a  ^  ^  I 

=  |a  6  c|'         y=\a  b  c|'  =|a  6  c  |' 

provided  |  a  6  c  |  =£  0. 

This  rule  for  finding  the  solution  of  (1)  is  due  to  Gabriel 
Cramer  (1760).  "We  proceed  to  prove  it. 

Assuming  that  equations  (1)  have  a  solution,  we  begin  by 
multiplying  them  respectively  by  +  Alt  —  A2)  +  A3,  i.e.  by  the 
signed  minors  of  al}  a?,  03  in  the  determinant  |  a  6  c  |.  Add- 
ing the  resulting  equations,  we  have 

-  bzA2  +  63^3)  y 


The  coefficient  of  x  is  the  evaluation  of  |  a  6  c  |  by  the  minors 
of  the  a's.  Similarly,  the  constant  term  is  |  k  b  c  \  ;  cf.  §  4, 

*  All  the  theorems  of  this  paragraph  have  been  proved  directly  from 
the  definition  of  a  determinant,  without  the  use  of  any  of  the  preceding 
theorems,  of  §§  4,  5,  6.  So  the  paragraph  could  be  inserted  immediately 
after  §  3.  Its  importance,  in  comparison  with  that  of  §§  4,  5,  6,  is  not, 
however,  sufficient  to  justify  this. 


382 


ANALYTIC   GEOMETRY 


Ex.  1.     The  coefficients  of  y  and  z  are  |  b  b  c\  and  |  c  6  c  |, 
and  these  determinants  are  zero  (Th.  4).     Consequently, 

|a  b  c\x=\k  b  c\. 
Since  we  are  assuming  that  |  a  b  c  \  •=£  0, 

\k  b  c\ 

/f  —  i . 

—  1 1 i 

1  a  b  c\ 

This  is  the  value  of  x,  as  given  by  (2).  Multiplying  the 
equations  (1)  respectively  by  —  B1}  -f  jB2,  —  J53  and  adding, 
we  obtain  the  value  of  y.  That  •  of  z  is  arrived  at  in  a  similar 
manner. 

What  we  have  proved  is  this :  If  the  equations  (1)  have  a 
solution,  it  is  given  by  formulas  (2).  It  follows,  then,  that 
equations  (1)  have  at  most  one  solution,  since  formulas  (2)  give 
unique  values  for  x,  y,  z. 

It  remains  to  show  that  these  values  of  x,  y,  z  actually  are  a 
solution,  i.e.  actually  satisfy  equations  (1)  in  all  cases.  This 
can  be  done  by  direct  substitution.  Setting  the  values  into 
the  first  of  equations  (1)  and  multiplying  through  by  |  a  b  c  |, 
we  have 

(3)     ^  |  fc  b  c\  +  bi\a  Jc  c  \  +  c^  \  a  b  Tc  \  —  Tc^  \a  b  c|  =  0. 

By  proper  rearrangement  of  columns  in  the  first  two  deter- 
minants (cf.  Ths.  6-8),  this  becomes 

Oi  |  6  c  fe  |  —  &!  |  a  c  A;  |  +  G!  |  a  b  k\—  &j  |  a  6  c  |  =  0. 
The  left-hand  side  here  is  the  evaluation  of  the  determinant 


&1 


«3 


by  the  minors  of  the  first  row.  But  this  determinant  is  zero, 
because  the  first  two  rows  are  identical.  Consequently,  (3)  is 
a  true  equation  and  the  values  of  x,  y,  z  given  by  (2)  satisfy 


DETERMINANTS  383 

the  first  of  the  equations  (1).     In  like  manner  it  can  be  shown 
that  they  satisfy  the  other  two  equations. 

This  completes  the  proof  that  equations  (1),  provided  |  a  b  c  \ 
^=  0,  have  a  unique  solution,  which  is  given  by  Cramer's  rule. 
Both  the  proof  and  the  rule  can  be  generalized  to  the  case  of 
any  number  of  simultaneous  linear  equations  in  the  same  num- 
ber of  unknowns.  We  state  the  result  in  general  form.  - 

THEOREM  10.  A  number  of  simultaneous  linear  equations  in 
the  same  number  of  unknowns,  for  which  the  determinant  of  the 
coefficients  of  the  unknowns  does  not  vanish,  has  one  and  only  one 
solution,  which  is  given  by  Cramer's  rule.* 

EXERCISES 

1.  Deduce  the  value  of  y  given  by  (2). 

2.  Prove  that  the  values  of  x,  y,  z   given   by    (2)    actually 
satisfy  the  third  of  equations  (1). 

3.  Give  Cramer's  rule  for  four  simultaneous  linear  equations 
in  four  unknowns.     First  write  down  the  equations  and  then 
the  formulas  analogous  to  formulas  (2).     No  proof  is  required. 

Solve  the  following  systems  of  simultaneous  equations. 
2x-    y  +  3z+    t=     6,  3y-4z+2£-  4  =  0, 


=  —  6,  2x         +3z-4£+  3  =  0, 

'    3x-2y  —    z  +  4*  =  -l,  -4x+2y 


=     8.  3x-4+2z  -  5  =  0. 


9.   Three  Equations  in  Two  Unknowns.     Compatibility.     The 

three  linear  equations, 

«!*  +  %  +  Cj  =  0, 

(1)  a^x  +  bfl  -f  c2  =  0, 

a&  +  batf  +  C3  =  0, 

*  In  case  the  determinant  of  the  coefficients  does  vanish,  the  facts  are 
more  complex.  For  two  equations  in  two  unknowns,  they  are  given  in 
the  footnote  on  p.  360.  For  a  treatment  of  the  general  case,  cf  .  Bdcher, 
Introduction  to  Higher  Algebra,  Ch.  IV. 


384  ANALYTIC   GEOMETRY 

in  the  two  unknowns  x,  y  are  said  to  be  compatible,  or  consist- 
ent, if  they  have  a  simultaneous  solution.  They  will,  in 
general,  be  incompatible,  since  a  solution  of  two  of  them  will 
not,  in  general,  satisfy  the  third.  It  is  important,  then,  to 
determine  the  condition  for  their  compatibility. 

Two  cases  arise,  according  as  the  minors  d,  C2,  C3  of  the 
elements  c:,  c2,  c3  in  the  determinant  |  a  b  c  \  are  not,  or  are, 
all  zero.  In  case  they  are  not  all  zero,  at  least  one  of  them 
must  be  different  from  zero.  Suppose  that  (73  =  |  al  b2\  is 
not  zero.  Then  the  first  two  of  the  equations  (1)  have,  by 
Th.  10,  one  and  only  one  solution,  namely  : 

(2)  x 


This  will  be  a  solution  of  the  third  equation  if  and  only  if 
«s  1  °i  GZ  |  —  63  tti  €2  1  +  c3  1  at  62  1  =  0, 

or,  since  the  left-hand  side  here  is  the  expansion  of  |  a  b  c  |  by 
the  minors  of  the  third  row,  if  and  only  if 

(3)  |  a  b  c|  =  0. 

Before   formulating   this   result   as   a   theorem,  we  give  a 
definition. 

DEFINITION.     The  numbers  al}  a2,  a3  and  b},  62,  53  are  pro- 
portional : 

ot!  :  a2  :  a3  =  &!  :  52  :  b3, 

if  and  only  if  there  exist  two  numbers  I  and  m,  not  both  zero,  such 
that 

(4)  Zaj  =  m&x,  Ia2  =  mb2,  las  =  mb3. 


If  &u  &2>  &3  are  a^  zero  an(i  we  ^ake  I  =  0  and  m  any  number 
^fc  0,  equations  (4)  are  satisfied  no  matter  what  values  ab  a2,  a3 
have.  In  other  words,  three  arbitrary  numbers  a1}  a?,  a3,  on 
the  one  hand,  and  0,  0,  0,  on  the  other,  are  always  proportional. 
In  particular,  a1?  a.2,  a3  may  also  all  be  zero. 

Suppose,  now,  that  bi,  62,  63  are  not  all  zero  and  let  blt  for 
example,  be  not  zero.  If,  then,  I  were  0,  we  should  have, 


DETERMINANTS  385 

from  the  first  of  equations  (4),  m  =  0 ;  but  I  =  0,  m  =  0  is  con- 
trary to  the  definition.  Consequently,  in  this  case,  I  cannot 
be  0.  Hence  we  can  divide  each  equation  through  by  I.  The 
result  is  the  equations 

di  =  kbly  a2  =  fc&2,  as  =  &&3> 

where  7c  has  the  value  m/l.  Conversely,  if  in  any  given  case 
there  exists  a  number  k,  zero  or  not  zero,  such  that  these  equa- 
tions hold,  then  a1}  a2,  a3  and  ftj,  &2,  b3  are  proportional.  For, 
the  equations  are  but  a  special  case  of  equations  (4),  when 
I  =  1  and  m  =  k.  We  have  thus  proved  the  following  theorem  : 
Ifbi,  62,  63  are  not  all  zero,  the  numbers  a1?  cu,  a3  and  bi,  62,  b3 
are  proportional  if  and  only  if  there  exists  a  number  k,  zero  or 
not  zero,  such  that 
(5)  •  ax  =  kbi,  a?  =  kb2,  a3  =  kb3. 

By  application  of  the  definition  it  is  easy  to  show  that  the 
minors  Cl}  C2,  C3  in  the  above  discussion  are  not  all  zero  when 
and  only  when  c^,  a2,  a3  and  6j,  62,  63  are  not  proportional ;  cf. 
Ex.  2.  The  foregoing  result  can  be  stated,  then,  as  follows  : 

THEOREM  11.     If,  in  the  equations  (1), 
aj :  a2  :  a3  =£  b^ :  b2  :  b3, 

the  three  equations  will  be  compatible  when  and  only  when  the 
determinant  of  their  coefficients  vanishes.  TJiey  then  have  one 
and  only  one  solution. 

The  case  in  which  C^  =  C2  =  C3  =  0  is  left  to  the  student ; 
cf.  Ex.  3. 

EXERCISES 

1.  Show  that  the  equations, 

aix  -f-  &!  =  0,         atfc  +  62  =  0,         al  =£  0,     o^  =£  0, 
are  consistent  if  and  only  if  |  a  6 1  =  0. 

2.  The  proportion  a± :  a2  :  a3  =  6X :  62  :  63  is  valid  if  and  only 
if  the  three  two-rowed  determinants,  which  are  formed  from 

the  array  » 

«i    a2     03 

bi     b2     b3 


386  ANALYTIC   GEOMETRY 

by  dropping  each  column  in  turn,  are  all   zero.     Prove   this 
theorem. 

3.  If  in  the  equations  (1)  a},  a2,  a3,  bl}  b2,  b3  are  not  all  zero 
and  ai}  a2,  a3  are  proportional  to  blf  b2,  63,  the  three  equations 
will  be  compatible  if  and  only  if  ci}  c2,  <%  are  proportional  to  ai} 
0%,  a3  and  bl}  b2,  bs.  They  then  have  infinitely  many  solutions. 
Prove  this  theorem. 

In  each  of  the  following  exercises  determine  whether  or  not 
the  given  system  of  simultaneous  equations  is  compatible.  If 
it  is,  find  the  solution. 


4.  3X-    y+5  =  o, 

5.  2x+    ?/-l  =  0,       3x  — 

6.  x-2y  +  3  =  0,  -3x  +  6y-9  =  0,       2x-  4^  +  6  =  0. 

7.  4x  +  2?/-l  =  0,       2x+    y- 


8.    THEOREM.     If,  in  the  equations 

aix  +  hy  +  CjZ  +  dj  =  0, 
d%  =  0, 
d3  =  0, 
+  d4  =*  0, 

the  four  minors  DI,  Dz,  D3,  -D4  of  the  elements  d1?  d2)  d^,  d4  in  the 
determinant  \  a  b  c  d  \  are  not  all  zero,  the  equations  are  com- 
patible when  and  only  when  \a  b  c  d  \  =  0.  TJtey  then  have  one 
and  only  one  solution.  Prove  this  theorem. 

Determine  in  each  case  if  the  four  given  equations  are  com- 
patible.    If  so,  what  is  the  common  solution  ? 

4:X-2y  +  2z  —  5  =  0, 
2x-    y+ 


, 
z  +  5  =  0,  6x  -  3y  +  3z  +  4  =  0, 

2  +  2  =  0.  -2x+    y-    2  +  2  =  0. 

11.  State  the  generalization  of  Theorem  11  and  the 
theorem  of  Ex.  8  for  the  case  of  n  -j-  1  linear  equations  in  n 
unknowns. 


DETERMINANTS  387 

10.   Homogeneous  Linear  Equations.     The  equations, 

«i*  +  bty  +  cjz  =  0, 
(1)  a«x  +  &<#  +  c2z  =  0, 

a3x  +  b3y  +  c3z  =  0, 

form  what  is  called  a  system  of  homogeneous  *  linear  equations. 
In  considering  them,  we  assume  that  not  all  the  coefficients 

ai>  °i)  '  •  *>  cs  are  zero- 
Let  £0,  y0,  20  be  a  simultaneous  solution  of  the  equations  (1). 
Then  kx0,  ky0,  kz0,  where  k  is  an  arbitrary  constant,  is  also  a 
solution  of  (1).     For,  if  these  values  are  substituted  for  x,  y,  z 
in  (1),  we  have 

k  (0,1X0  +  6iy0  +  Ci«b)=  0, 
k  (a,,x0  +  &2?/0  +  c2Zo)  =  0, 
fc  fa*k  +  b3y0  +  c320)  =  0. 

The  three  parentheses  in  these  equations  all  have  the  value 
zero,  since  x0,  y0,  z0  is  a  solution  of  (1).  Hence  the  equations 
are  true,  q.  e.  d. 

This  proof  is  applicable  to  the  general  case  of  n  homogene- 
ous linear  equations  in  n  unknowns.  Hence  we  can  state  the 
following  theorem  : 

THEOREM  12.  If  x0,  y0,  z0,  ••-,  t0  is  a  simultaneous  solution 
of  n  homogeneous  linear  equations  in  the  n  unknowns  x,  y,  z,  •  •  •, 
t,  then  kx0,  ky0,  kz0)  ••-,  kt0,  where  k  is  an  arbitrary  constant,  is 
also  a  solution. 

An  obvious  solution  of  the  equations  (1)  is  0,  0,  0.  This 
is  the  only  solution,  if  the  determinant  |  a  b  c  |  is  not  0.  For, 
equations  (1)  are  a  special  form  of  equations  (1),  §  8,  when 
ki  =  k%  =  A;3  =  0.  If  |  a  b  c  \  =£  0,  the  latter  equations  have, 
by  Th.  10,  just  one  solution,  given  by  formulas  (2),  §  8.  But 
this  solution  is  0,  0,  0,  since  each  of  the  determinants  in  the 
numerators  in  (2)  now  contains  a  column  of  zeros. 
This  result  is  also  general : 

THEOREM  13.  If  the  determinant  of  the  coefficients  of  n  homo- 
geneous linear  equations  in  the  n  unknowns  x,  y,  z,  ••  •,  t  does 

*  Cf.  p.  348. 


388  ANALYTIC   GEOMETRY 

not  vanish,  the  only  simultaneous  solution  of  the  equations  is 
a  =  0,  y  =  0,  2  =  0,  ••-,*  =  0. 

If,  then,  the  equations  are  to  have  a  solution  other  than  the 
obvious  solution  0,  0,  0,  •  •  •,  0,  it  is  necessary  that  the  determi- 
nant of  the  coefficients  vanish.  It  can  be  shown,  conversely, 
that  if  this  determinant  does  vanish  the  equations  will  have 
solutions  other  than  the  obvious  solution.  That  is,  the  follow- 
ing theorem  is  true. 

THEOREM  14.  A  system  of  n  homogeneous  linear  equations  in 
n  unknowns  has  a  solution  other  than  the  obvious  solution,  0,  0, 
0,  •••,  0,  if  and  only  if  the  determinant  of  the  coefficients  vanishes. 

To  complete  the  proof  of  this  theorem  in  the  case  of  equa- 
tions (1),  we  must  show  that,  if  |  a  6  c  |  =  0,  the  equations  have 
solutions  other  than  0,  0,  0.  This  we  shall  do  by  actually  ex- 
hibiting such  solutions. 

By  hypothesis,  \a  b  c|  =  0.  Then  two  cases  arise,  as 
follows  : 

Case  1.  Not  all  the  minors  in  \a  b  c\  are  zero.  In  this  case 
at  least  one  minor  in  |  a  b  c  \  does  not  vanish.  Suppose  that 
the  minor,  |a]  b.2\,  in  the  upper  left-hand  corner  is  not  zero.* 
We  proceed,  then,  to  show  that  equations  (1)  have  a  solution 
x,  y,  z,  in  which  z  =  1 ;  i.e.  that  the  equations, 

a^x  +  biy  +  G!  =  0, 
(2)  a2x  +  b2y  +  c2  =  0, 

«s«  +  bsy  +  c3  =  0, 

obtained  from  the  equations  (1)  by  setting  2  =  1,  have  a  simul- 
taneous solution  for  x  and  y. 

Since,  by  hypothesis,  \a  b  c|=0  and  |a!  62|  =£  0,  equations 
(2)  have,  according  to  Th.  11,  just  one  solution,  that  given  by 
formulas  (2),  §  9,  namely : 


(3) 

*  If  in  any  particular  case  this  minor  were  zero,  the  equations  and  the 
terms  in  them  could  be  rearranged,  so  that  the  minor  would  not  be  zero. 


DETERMINANTS  389 

Consequently,  equations  (1)  have  the  solution 

/  »  \ 

(4)  x 

and  hence,  by  Th.  12,  the  solution 

(5)  x=\bj,  c.2|,         2/  =  -|ai  c2|,         z=|ai  68|, 
or,  finally,  again  by  Th.  12,  the  solutions 

(6)  x  =  Jc\bt  c2|,         y  =  —  k\a>i  c%\,        z  =  k\a,i  &2|. 

There  are  infinitely  many  solutions  given  by  (6),  since  k 
may  have  any  value.  Inasmuch  as  |  c^  &2  1  ^  0,  only  one  of 
these  solutions  is  the  solution  0,  0,  0,  namely  the  one  for  which 
A;  =  0.  Hence  the  theorem  is  proved  in  this  case. 

Furthermore,  (6)  gives  all  the  solutions  of  (1).  To  prove 
this,  let  x0,  y0,  z0  be  an  arbitrary  solution  of  (1).  If  z0  =  0, 
then  x0  =  y0  =  0,  since  for  z  =  0  the  first  two  of  equations  (1) 
become 

a-fl  +  b$  =  0,         a^x  +  b^  =  0, 

and  the  only  solution  of  these  equations  is  0,  0,  because 
|aj.  ^I^O;  cf.  Th.  13.  If  z0^Q,  then  XQ/ZO,  y0/zQ,  1  is  a  solu- 
tion of  (1)  and  XQ/ZQ,  yo/zo  is  therefore  a  solution  of  (2).  But 
the  only  solution  of  (2)  is  given  by  formulas  (3).  Hence  it 
follows  that 


z0 
or  that 

a0  =  fc|6;  C2J,         2/o  =  —  fc|«i  c2  1,         20  =  A;la1&2|> 

where  A;  has  a  definite  value,  not  zero. 

We  may  state  the  final  result  by  saying  that  every  solution 
of  (1)  is  proportional  to  the  solution  (5),  meaning,  thereby,  that 
it  is  given  by  equations  of  the  form  (6)  ;  cf.  §  9,  eq.  (5). 

Case  2.  All  the  minors  in  \  a  b  c\  are  zero.  -In  this  case  it 
follows,  by  §  9,  Ex.  2,  that 

aj  :  6X  :  G!  =  a2  :  62  :  Ca  =  a3  :  63  :  C3. 


390  ANALYTIC   GEOMETRY 

This  means  that  the  left-hand  sides  of  equations  (1)  are  pro- 
portional to  one  another.  Consequently,  all  the  solutions  of 
one  of  the  equations  are  solutions  of  the  other  two,  and  hence 
are  all  the  solutions  of  the  system  (1). 

The  equation  thus  singled  out  must  be  one  in  which  the 
three  coefficients  are  not  all  zero.  This  is  true  of  at  least  one 
of  the  equations  (1),  since,  by  hypothesis,  not  all  the  coefficients 
in  (1)  are  zero.  Let  it  be  true  of,  say,  the  first  equation : 

(7)  a&  -f  b$  +  CjZ  =  0, 

and  let  at,  for  example,  be  not  zero. 

In  solving  (7),  the  values  of  y  and  z  can  be  chosen  at 
pleasure :  y  =  k,  z  =  l,  and  the  value  of  x  is  then  determined. 
Consequently,  all  the  solutions  of  (7),  and  hence  of  (1),  are 
given  by 

(8)  a?  =  -fc^-Z^-,         y  =  k,        z  =  l. 

«!  tti 

Here  there  are  two  arbitrary  constants,  k  and  I.  We  say,  then, 
that  the  equations  (1)  have  a  two-parameter  family  of  solutions 
in  this  case;  and,  in  distinction,  a  one-parameter  family  of 
solutions  in  Case  1. 

The  proof  of  Theorem  14,  for  n  =  3,  is  now  complete.  In 
the  general  case  the  facts  and,  consequently,  the  proof  are 
much  more  complicated.*  See  p.  403,  footnote. 

EXERCISES 

1.  Prove  Theorem  14  for  the  case  n  =  2. 

2.  Prove  the  Theorem:   If  xi}  yv,  — ,  ^  and  x2,  y^  •••,  t2  are 
two  simultaneous  solutions  of  n  homogeneous  linear  equations  in 
the  n  unknowns  x,  y,  z,  ••-,  t,  then  xv  +  x%,  yl  +  y2,  •••,  ti  +  £2  *« 
also  a  solution.     Take  first  n  =  3. 

3.  (Continuation  of  Ex.  2.)     Show,  further,  that  kxl  +  lxz, 
ty\  +  ^2/2>  "•?  ^i  +  ^2  ls  a  solution. 

*  Cf.  Bdcher,  Introduction  to  Higher  Algebra,  Ch.  IV. 


DETERMINANTS  391 

Solve  the  following  systems  of  simultaneous  equations,  ob- 
taining all  the  solutions  in  each  case. 

4.  ±x-2y=0,         -6x  +  3y=Q. 

5.  3x+5y+8z=Q,        <lx—y+z=0,  x+2y—2z=0. 

6.  3x+2y-2z=Q,        2x+3y—z  =  Q,  8x+7y—5z—0. 

7.  x— 2y+3z=0,       —  3x+6y—9z=Q,  2x-4y+6z=0. 


II.     APPLICATIONS 

11.  The  Straight  Line.  Equation  of  the  Line  through  Two 
Points,  in  Determinant  Form.  Let  (x,  y)  be  an  arbitrary  point 
on  the  line  determined  by  the  two  points  (xl}  y^  (x«,  y2),  and  let 

(1)  Ax  +  By  +0=0 

be  the  equation  of  the  line.     Since  (x,  y),  (xlf  yi),  (#2,  yz)  lie 
on  the  line,  we  must  have 

Ax  +  By  +(7  =  0, 

(2)  Ax1  +  By1+C=0, 


These  equations  are  linear  and  homogeneous  in  the  three  un- 
knowns A,  B,  C.  They  have  a  solution  for  A,  B,  C  other  than 
the  obvious  solution  0,  0,  0,  inasmuch  as  there  is  a  line  (1)  on 
which  the  three  points  (x,  y),  (xl}  y^),  (0^,2/2)  lie-  Consequently, 
by  Th.  14,  the  determinant  of  the  coefficients  vanishes. 

In  other  words,  every  point  (x,  'y}  or,  on  dropping  the  dashes, 
every  point  (#,  y)  on  the  line  satisfies  the  equation 


(3) 


x      y 


=  o. 


By  a  careful  retracing  of  the  steps,  it  can  be  shown,  con- 
versely, that  every  point  (x,  y)  satisfying  (3)  lies  on  the  line. 
It  would  follow,  then,  that  (3)  is  the  equation  of  the  line.  We 
shall  adopt,  however,  a  quite  different  method  to  prove  this, 


392 


ANALYTIC   GEOMETRY 


regarding  the  foregoing  work  as  primarily  of  value  in  furnish- 
ing us  equation  (3). 

To  show  that  equation  (3)  represents  the  line  through 
(xi>  2/i)>  fat  2/a)>  develop  the  determinant  by  the  minors  of  the 
first  row.  Equation  (3)  then  takes  on  the  usual  form  (1)  of  a 
linear  equation  in  x  and  y ;  moreover,  the  values  obtained  for 
A  and  B : 

A  =  2/i  —  2/2  B  =  x2  —  x1, 

are   not  both   zero,  since  the  given  points  do  not  coincide. 
Consequently,  (3)  represents  some  straight  line. 

This  line  is  the  required  line,  if  the  coordinates  of  the  given 
points  satisfy  (3).  They  do,  for,  if  we  replace  x,  y  in  the  de- 
terminant by  a/1}  2/1,  or  by  0%,  y2,  two  rows  of  the  determinant 
will  be  identical  and  hence  the  value  of  the  determinant  will 
be  zero. 

Three  Points  on  a  Line.  Let  the  three  points,  which  we 
assume  are  distinct,  be  (xl}  2/1),  (&,  2/2)?  (%>  2/s)-  The  equation 
of  the  line  through  the  second  and  third  is,  according  to  (3), 

x     y     I 

(4)  x2    2/2     1=0. 

^        2/3        1 

The  first  point  lies  on  this  line  if  and  only  if  (xt,  yt)  -satisfies 
(4),  i.e.  if  and  only  if 


(5) 


2/2 
2/3 


=  0. 


This  result  we  state  as  follows  : 

THEOREM  15.     The  three  points  (x1}  y^,  (x.2,  3/2),  (#3>  2/s)  o,re 
collinear,  if  and  only  if  the  determinant  in  (5)  vanishes. 

Three  Lines  through  a  Point.     Consider  the  three  distinct 
lines 


(6) 


DETERMINANTS  393 

They  are  parallel,  by  Ch.  II,  §  10,  Th.  3,  if  and  only  if       . 

AI'.  B!  =  A2:  B2  =  As  .  B3, 

i.e.  if  and  only  if 

(7)  AI  :  A2  :  As  =  Bl :  B2  :  Bs. 

Suppose,  now,  that  the  three  lines  go  through  a  point. 
This  means,  analytically,  that  the  equations  (6)  have  a  common 
solution  for  x,  y,  i.e.  are  compatible.  Hence,  it  follows,  by 
Th.  11,  since  in  this  case  (7)  cannot  hold,  that  \ABC\  =  0. 

The  determinant  \  A  B  O\  vanishes  also  when  the  three 
lines  are  parallel,  since  then  (7)  is  valid  and  the  first  two 
columns  in  the  determinant  are  proportional. 

Conversely,  if  |  A  B  C  \  =  0,  the  lines  (6)  are  parallel  or 
concurrent.  For,  if  the  determinant  vanishes  by  virtue  of  the 
first  two  columns  being  proportional,  (7)  holds  and  the  lines  are 
parallel.  On  the  other  hand,  if  (7)  does  not  hold,  equations 
(6),  by  Th.  11,  are  compatible  and  this  means,  geometrically, 
that  the  three  lines  have  a  point  in  common. 

We  have  thus  proved  the  theorem : 

THEOREM  16.  The  three  lines  (6)  are  concurrent  or  parallel 
if  and  only  if  the  determinant  of  their  coefficients  vanishes: 

\A  B  C  |  =  0. 

EXERCISES 

Find  the  equations  of  the  following  lines  in  determinant 
form. 

1.  The  line  through   (xlf  y^  with  intercept  b  on  the  axis 
of  y- 

2.  The  line  with  intercepts  a  and  b. 

Find,  in  determinant  form,  the  equations  of  the  lines  re- 
quired in  the  following  exercises  of  Chapter  II.  Reduce  the 
equation  each  time  to  the  usual  form. 

3.  Ex.  1,  §  1.      4.  Ex.  2,  §  1.        5.  Ex.  4,  §  1.       6.  Ex.  6,  §  1. 
7.  Ex.  7,  §  1.      8.  Ex.  10,  §  1.      9.  Ex.  1,  §  5.     10.  Ex.  3,  §  5. 


394 


By  the  method  of  this  paragraph,  do  the  following  exercises 
at  the  end  of  Chapter  III  concerning  three  lines  through  a 
point  or  three  points  on  a  line. 

11.  Ex.  1.  12.    Ex.  2.  13.   Ex.  3. 
14.   Ex.  4.                       15.    Ex.  5.                      16.    Ex.  6. 

Are  the  lines  given  in  the  following  exercises  concurrent  ? 
parallel  ? 

17.   Ex.  4,  §  9.  18.   Ex.  5,  §  9.  19.    Ex.  7,  §  9. 

12.  The  Circle  and  the  Conies.     Equation  of  the  Circle  through 
Three  Points.     If   the   three   points  (xl}  yi),  (x2,  y^),  fa,  y^, 
which  we  assume  are  not  collinear,  lie  on  the  circle 

(1) 

it  follows  that 


(2) 


Cy3  +  D  =  0. 


(3) 


=  o. 


In  (1)  and  (2)  we  have  four  homogeneous  linear  equations 
in  the  four  unknowns  A,  B,  C,  Z>,  which  have  a  solution  other 
than  the  obvious  solution,  0,  0,  0,  0.  Consequently,  by  Th.  14, 

x     y      I 

*i   y\   i 

X2      7/2       1 

a*   y3    i 

Equation  (3)  is  the  equation  of  the  circle  through  the  three 
given  points.  For,  if  we  develop  the  determinant  in  (3)  by 
the  minors  of  the  first  row,  we  obtain  an  equation  of  the  form 
(1),  where 

y\ 

A—   xz    2/2     1^=0, 


since  the  three  points  were  assumed   non-collinear  (Th.  15). 
Consequently,  equation  (3)  represents  a  circle,  or  a  point,  or 


DETERMINANTS 


395 


it  has  no  locus ;  cf.  Ch.  IV,  §  2.  That  it  represents  a  circle, 
and,  in  particular,  the  required  circle,  is  clear  since  the  coor- 
dinates of  each  of  the  three  points  satisfy  it. 

Condition  that  Four  Points  Lie  on  a  Circle. 
THEOREM  17.     The  four  points  fa,  y^,  fa,  y2)>  fa,  2/s),  fa,  y<), 
oftvhich  we  assume  no  three  collinear,  lie  on  a  circle  if  and  only  if 


(4) 


2/i2 


=  0. 


-t 

7/1        _L 

2/2    i 

I  2  "1 

~r  2/3      xz     2/s     -1- 
+  2/42     z4     2/4     1 
The  proof  is  left  to  the  student. 

Conic  through  Five  Points.  The  general  equation  of  the 
straight  line  (1),  §  11  contains  three  constants,  A,  B,  C,  enter- 
ing homogeneously  —  one  in  each  term  —  and  we  can  always 
pass  just  one  line  through  two  points.  Also,  the  general 
equation  (1)  of  the  circle  contains  four  homogeneous  constants 
A,  B,  C,  D,  and  through  three  points  (non-collinear)  we  can 
always  pass  just  one  circle. 

The  general  equation  of  a  conic, 

contains  six  homogeneous  constants,  and  accordingly  we 
should  expect  that  through  FIVE  points  we  can,  in  general,  pass 
just  one  conic. 

We  prove  this  by  writing  down  the  equation  of  the  conic 
through  the  five  points  fa,  y^,  fa,  y2),  fa,  y,),  fa,  y4),  fa,  y,). 

Proceeding  as  in  the  cases  of  the  straight  line  and  circle, 
we  find  as  the  probable  equation  : 


(6) 


x2       xy      y2     x     y 


1 

•4 

2/2  1 

2/3  1 

2/4  1 

2/5  1 


=  0. 


396  ANALYTIC   GEOMETRY 

When  the  determinant  is  developed  by  the  minors  of  the 
first  row,  equation  (6)  takes  on  the  form  (5).  -Two  cases  then 
arise,  according  as  the  values  obtained  for  A,  B,  C  are  not,  or 
are,  all  zero. 

Case  1.  '  A,  B,  C  not  all  zero.  In  this  case  it  follows  that 
equation  (6)  represents  some  conic,  in  particular,  a  conic 
through  the  five  given  points,  since  it  is  clear  that  the  coordi- 
nates of  each  of  the  points  satisfy  the  equation. 

We  state,  without  proof,  that  this  case  occurs  unless  four, 
or  all  five,  of  the  given  points  are  collinear. 

If  no  three  of  the  points  are  collinear  the  conic  just  found 
must  be  non-degenerate.  It  is  the  only  conic  through  the 
five  points.  For,  if  there  were  a  second  conic  through  them, 
the  two  conies  (both  non-degenerate)  would  intersect  in  five 
points,  and  this  is  impossible.* 

If  three  of  the  points  are  collinear,  the  conic  found  must 
be  degenerate ;  f  in  particular  it  must  consist  of  two  straight 
lines   (Fig.  1).     Clearly,  these   lines 
are  uniquely  determined  by  the  five 
points  and  hence  so  is  the  conic. 
FlQ  j  The  results  of  this  case  we  formu- 

late as  a  theorem : 

THEOKEM  18.  Through  Jive  points,  no  four  of  which  are 
collinear,  there  passes  one  and  only  one  conic.  If  three  of  the 
points  are  collinear,  the  conic  is  degenerate ;  otherwise,  it  is  non- 
degenerate. 

Case  2.  A  =  B  =  C=0.  Then  D  =  E  =  F=  0  also,  and 
equation  (6)  reduces  to  the  trivial  equation :  0  =  0.  Stated 
without  proof. 

This  case  occurs  if  at  least  four  of  the  five  points  are  col- 

*  That  two  non-degenerate  conies  cannot  intersect  in  more  than  four 
points  is  geometrically  evident ;  an  analytical  proof  is  beyond  the  scope  of 
this  book. 

I  If  it  were  non-degenerate,  we  should  have  a  non-degenerate  conic 
intersected  by  a  line  in  three  points  —  an  impossibility. 


DETERMINANTS  397 

linear.  If  just  four  are  collinear,  there  are  infinitely  many 
degenerate  conies  through  the  five  points,  each  consisting  of 
the  line  of  the  four  points  and  some  line  through  the  fifth. 

If  all  five  points  are  collinear,  their  line,  taken  with  any 
line  in  the  plane,  forms  a  degenerate  conic  through  them,  so 
that  here,  too,  there  are  infinitely  many  degenerate  conies 
through  the  five  points.* 

Parabolas  through  Four  Points.  Demanding  that  the  conic 
defined  by  equation  (5)  be  a  parabola  puts  one  condition  on 
the  coefficients  in  (5),  namely, 

(7)  B2-  ±AC=0. 

Consequently,  we  cannot  prescribe  more  than  four  points 
through  which  a  parabola  must'  pass. 

Let  (0,  0),  (1,  1),  (-  1,  1),  (3,  9)  be  the  four  points.     Then 

F=Q, 

A+      B  +      C+    D  +    E+F=0, 
A-       B+      C-    D+    E  +  F=0, 


To  solve  equations  (7)  and  (8)  simultaneously,  find  the  values 
of  D,  E,  F  in  terms  of  A,  B,  C  from  the  first  three  of  equa- 
tions (8)  : 

(9)  D  =  -B,        E  =  -A-C,        F=0, 

and  substitute  them  in  the  fourth  equation.     The  result  is 

(10)  B  =  -  3  C. 
Hence  (7)  becomes 


and  (7=0         or         C  =  $A. 

Prom  equations  (9)  and  (10)  we  have,  then  : 

(7=0,          £  =  0,  D  =  0,          E  =  -A,          F=Q, 

orC=$A,       B  =  -±A,       D  =  ±A,       E  =  -^A,      F=0. 

*  There  is  a  one-parameter  family  of  degenerate  conies  in  the  first 
case,  a  two-parameter  family  in  the  second;  cf.  p.  390.  Can  the  student 
explain  why  ? 


398  ANALYTIC   GEOMETRY 

Setting  A  =  1  in  the  first  case  and  A  =  9  in  the  second,  we 
find  as  the  resulting  equations 

x2  -  y  =  0, 
9x2  —  I2xy  +  4  #2  +  12x-  13y  =  0. 


There  are,  then,  two  parabolas  through  the  four  given  points. 
We  state  without  proof  that  this  is,  in  general,  true.  Of 
course,  one  or  both  of  the  parabolas  may  be  degenerate,  and 
for  special  positions  of  the  four  points  the  two  may  coincide. 
Finally,  if  the  four  points  are  collinear,  there  are  an  infinite 
number  of  degenerate  parabolas  through  them. 

EXERCISES 

1.  State  and  prove  the  theorem  giving  the  condition  that 
six  points,  no  four  of  which  are  collinear,  lie  on  a  (non-degen- 
erate or  degenerate)  conic.     If  four  or  more  of  the  points  are 
collinear,  is  there  a  conic  through  the  six  ? 

Find,  in  determinant  form,  the  equations  of  the  circles  re- 
quired in  the  following  exercises  of  Chapter  IV.  Reduce  the 
equation  each  time  to  the  usual  form. 

2.  Ex.  1,  §  4.  3.   Ex.  2,  §  4.  4.   Ex.  3,  §  4. 

In  each  of  the  following  exercises  determine  whether  or  not 
the  four  given  points  lie  on  a  circle. 

5.  (0,0),  (3,0),  (0,1),  (2,-  1). 

6.  (2,0),  (-3,0),  (0,4),  (-1,4). 


7.    (a,  0),  (b,  0),  (0,  c),  (0, 
V 


Find,  in  each  exercise  that  follows,  the  equation  of  the  conic 
through  the  given  five  points.     Is  the  conic  non-degenerate  ? 

8.  (0,  0),  (2,  0),  (0,  2),  (5,  2),  (2,  5). 

Ans.    2z2  —  3xy  +  2y*  —  4a-  4y  =  0. 

9.  (1,0),  (-1,0),  (0,1),  (0,  -1),  (1,1). 


DETERMINANTS 


399 


10.  (1,  -  1),  (1,  1),  (3,  11),  (-3,  - 11),  (5,  19). 

Ans.    15  x2—  y2  =  14. 

11.  (0,0),  (2,1),  (3, -4),  (0,2),  (-2,0). 

12.  (1,  2),  (0,  1),  (6,  - 1),  (-  1,  -  2),  (3,  0). 

Find  the  equations  of  the  parabolas  through  each  of  the 
following  sets  of  four  points.  Are  they  degenerate  or  not  ? 

13.  (0,0),  (1,1),  (1,  -1),  (4,2). 

Ans.   yz  —  x  —  0 ;  (x  —  y)(x  —  y  —  2)  =  0. 

14.  (0,  0),  (3,  1),  (1,  3),  (6,  3). 

15.  (2,0),  (0,1),  (-1,1),  (5,  -2). 

16.  (2,  1),  (7,  0),  (4,  3),  (5,  -2). 

In  each  of  the  following  exercises  determine  whether  or  not 
the  six  given  points  lie  on  a  conic.  If  they  do,  find  if  the 
conic  is  degenerate. 

17.  (0,  0),  (1,  -  1),  (1,  3),  (5,  5),  (2,  4),  (6,  3). 

18.  (-  1,  - 1),  (0,  2),  (-  1,  0),  (5,  2),  (0,  - 1),  (9,  5). 

19.  (0,  1),  (1,  0),  (1,  -  1),  (3,  1),  (- 1,  3),  (-3,  -2). 

20.  (0,  0),  (2,  0),  (-1,  1),  (3,  1),  (5,  - 1),  (-  4,  2). 


EXERCISES  ON   CHAPTER  XVI 


Evaluate  each  of  the  following  determinants,  expressing  the 
result,  if  it  is  different  from  zero,  in  factored  form. 


1. 


4. 


a     a2 
b     62 


2. 


a  +  6 

b  +c 
c  +  a 


ab  c 
be  a 
ca  b 


3. 


0  -a  -b 
a  0  —  c 
b  c  0 


a  —  b 
b  —  c 
c  —  a 


Ans.  to  Ex.  1.    (a  —  6)(6  —  c)(c  —  a). 
Ans.  to  Ex.  2.    —  (a  —  6) (6  —  c)(c  -  a)  (a  +  b  +  c). 

1     a     a2 

b3 
c3 


a  +  0 

&4-c 
c  +a 


5. 


6  62 
c  c2 
d  d2 


400 


ANALYTIC   GEOMETRY 


Prove  that  the  following  determinants  have  the  value  zero. 

&!       2  C&!  +  3  &! 


6. 


7. 


o<>     62     2  a2  +  3  b 
a3     63     2  a3  +  3  b, 

1  3  e»!  —  &!  +  2     at 
—  1     3  a2  —  62  —  2     a2 

2  3  a3  —  63  +  4    a3 
1     3  a4  —  &4  +  2     a4 


&! 

62 

&3 
64 

Definition.  Let^,  p2,  —,pn,  g\,  g2,  •••,  gn,  and  r^,  r2,  ••-,  rn  be 
three  columns  (or  rows)  of  a  determinant.  The  third  is  said 
to  be  a  linear  combination  of  the  first  two,  if  two  numbers, 
Jc  and  I,  exist  such  that 


In  the  determinant  of  Ex.  6,  for  example,  the  third  column  is 
a  linear  combination  of  the  first  two ;  and  in  that  of  Ex.  7  the 
second  column  is  a  linear  combination  of  the  third,  fourth,  and 
first. 

8.  THEOREM.  If  one  column,  or  row,  of  a  determinant  is  a 
linear  combination  of  two  others,  the  value  of  the  determinant  is 
zero.  Prove  this  theorem.  How  can  it  be  extended  ? 

Solve  the  following  equations  for  x. 


9. 


10. 


Determine  k  so  that  the  following  equations  have  solutions 
other  than  0,  0,  0 ;  then  find  the  solutions. 


x 

+  1 

4 

2 

X 

-9 

5 

— 

3 

=  0. 

X 

-1 

— 

1 

1 

x  —  5 

2 

— 

1 

3 

6 

-3x 

4 

2 

7 

x 

+  4 

3 

-1 

=  0. 

x 

6 

2 

4 

DETERMINANTS 


z  =  0, 


11. 


kx-    y 

12.   4  03  —  2y 

6x-3y 


401 
z  =  0, 


Determine  k  so  that  the  following  equations  are  compatible. 
Find  the  common  solution  in  each  case. 


2x  —  3y-  k  =  0,  x  — 

13.    kx  —    y  —  k  =  0,  14.    kx  — 

kx  +3y  —  5  =  0.  z-f 

Find  all  the  solutions  of  the  following  equations. 
x  +  y  —z  =  0, 


+  5    =0, 
—  1    =0. 


15. 


16. 


17.    Show  that  all  the  solutions  of  the  equations, 

n&  =  0, 


are  given  by       x:y:z  =  |mn|:|nZ|:|Zm|, 

provided  not  all  three  of  the  determinants  on  the  right  are  zero. 

APPLICATIONS 

18.  Show  that  the  area  of  the  triangle  with  vertices  at  the 
points  (a?!,  y^,  (xz,  y8),  (x,,  y3)  is 

2/i 
2/2 
2/3 

19.  Prove  that  the  equation  of  the  line  of  slope  X  through 
the  point  (xl}  y^)  can  be  written  in  the  form 

x     y      I 

»!    2/1     1=0. 

1X0 

20.  Show  that  every  equation  of  the  form 

x      y      1 

a3  =  0, 


402 


ANALYTIC   GEOMETRY 


where  the  minors  of  x  and  y  are  not  both  zero,-  represents  a 
straight  line. 

21.    Show  that  the  points  (xlf  y^,  (a^,  y2)  are  collinear  with 
the  origin  when  and  only  when 


*a     2/i 

«2       2/2 


=  0. 


22.  Prove  that  the  distinct  lines  LI,  L2  of  Ch.  II,  §  10,  are 
parallel  if  and  only  if 

A\  J=0'  '"• 

23.  Show  that  the  lines  LI,  L2  of  Ch.  II,  §  10,  are  identical, 
if  and  only  if  the  three  two-rowed  determinants,  which  are 
formed  from  the  array 

Al    B,     C, 


by  dropping  each  column  in  turn,  are  all  zero. 

24.    Show  that  the  discriminant,  A,  of  the  quadratic  equation 

Ax*  +  Bx+C=Q, 

(cf .  Ch.  IX,  §  5)  can  be  written  in  the  form 

2A       B 
B    2C' 


A=  - 


25.  Show  that  the  discriminant,  A,  of  the  general  equation 
of  the  second  degree  in  x  and  y  (cf.  Ch.  XII,    §  4)    can   be 
written  as 

2A       B       D 

B    2C      E. 
D       E    2F 

26.  Prove  that  the  polars  of  all  points  (having  polars)  with 
respect  to  a  degenerate  conic  are  concurrent  or  parallel. 

Suggestion.     The  conic  can  be  represented  either  by 

by*  =  0  or  by  y2  =  c. 


DETERMINANTS 


403 


27.  By  applying  Ex.  26,  show  that  the  general  equation  of 
the  second  degree  represents  a  degenerate  conic  when  and 
only  when  its  discriminant,  as  given  by  the  determinant  in 
Ex.  25,  vanishes. 

Suggestion.  Demand  that  the  polars  of  three  non-collinear 
points,  as  (0,  0),  (1,  0),  (0,  1),  be  concurrent  or  parallel.  The 
equation  of  the  polar  of  (xlt  y\)  is  that  of  Ex.  2,  p.  188. 

28.  Prove  that,  if  the  general  equation  of  the  second  degree 
represents  a  non-degenerate  conic,  the  line  ax  +  by  +  c  =  0  will 
be  tangent  to  the  conic  if  and  only  if 


2A  B  D  a 

B  2C  E  b 

D  E  2F  c 

a  b  c  0 


=  0. 


Suggestion.  Apply  the  second  method  of  Ch.  IX,  §  5.  The 
equation  of  the  tangent  at  (xl}  yt)  is  given  by  Ex.  2,  p.  188. 

Note  to  p.  390.  Theorem  14  leads  to  an  important  result  concerning 
the  compatibility  (cf.  §  9)  of  equations  (2),  p.  388,  namely  : 

THEOREM.  If  equations  (2)  are  compatible,  the  determinant  of  their 
coefficients  vanishes. 

For,  if  equations  (2)  have  a  solution,  x$,  y0,  then  equations  (1)  have 
a  solution,  xo,  yo,  1,  not  the  obvious  solution,  0,  0,  0.  Consequently,  by 
Th.  14,  |  a  6  c  |  =  0. 

The  extension  of  the  theorem  and  the  proof  to  the  equations  of  §  9, 
Ex.  8,  and  to  the  general  case  of  §  9,  Ex.  11,  is  immediate. 

The  determinant  of  the  coefficients  of  the  equations  of  §  9,  Ex.  7, 
vanishes;  the  equations  are,  however,  incompatible,  —  they  represent 
three  parallel  lines.  In  other  words,  the  converse  of  the  theorem  is  not 
true  ;  cf .  Th.  11. 


CHAPTER   XVII 
PROJECTIONS.     COORDINATES 

1.  Directed  Line-Segments.  In  the  Introduction  to  Plane 
Analytic  Geometry  directed  line-segments  on  a  line  L  were 
denned  and  discussed.  Since  L  might  be  situated  anywhere  in 
space,  the  theory  there  developed  holds  equally  well  for  the 
geometry  of  space.  The  student  should  review  the  details  of 
this  theory.  Of  the  formulas,  let  him  recall  in  particular  the 
relation, 


(1)         MMl  +  MM  +  .-  +  Mn_zMn_i  +  Mn^N  =  MN, 

which  holds  for  any  n  -f-  1  points,  M,  MI,  M2,  •••  ,  Mn_i}  N, 
lying  on  L. 

2.  Projection  of  a  Broken  Line.  Given  a  point  P  and  a  line 
Z<  in  space.  The  projection  of  P  on  L  is  denned  as  the  foot,  M} 
of  the  perpendicular  dropped  from  P 
on  L,  or  as  the  point  M  in  which  the 
plane  p,  passing  through  P  perpen- 
dicular to  L,  meets  L.  If  P  lies  on 
L,  it  is  its  own  projection  on  L. 

Let  PQ  be  any  directed  line-seg- 
ment in  space  and  let  M  and  N  be 
the  projections  of  P  and  Q  on  L' 
The  projection  of  the  directed  line- 
segment  PQ  on  L  is  denned  as  the 
directed  line-segment  MN. 

405 


FIG. 


406  ANALYTIC   GEOMETRY 

If  p  and  q  are  the  planes  through  P  and  Q  perpendicular  to 
L,  the  projection,  MN,  of  PQ  on  L  is  equal  to  the  directed  line- 
segment  intercepted  by  the  planes  p  and  q  on  any  parallel  to  L. 
For  example,  it  is  equal  to  the  directed  line-segment  PR  in 
Fig.  1  * 

Consider  a  broken  line  joining  P  to  Q  and  consisting  of  the 
directed  line-segments  PP1?  PiP2,  •••,  Pn_iQ,  which  do  not 
necessarily  lie  in  a  plane.  The  sum  of  the  projections  of  these 
directed  line-segments  is 

MMl 


By  (1),  §  1,  this  sum  is  equal  to  MN,  i.e.  to  the  projection  on 
L  of  the  directed  line-segment  PQ. 

Thus  Theorem  1  of  the  Introduction,  §  3,  is  extended  to  the 
geometry  of  space  : 

THEOREM.  The  sum,  of  the  projections,  on  any  line  L  of  space, 
of  the  directed  line-segments,  PP],  P\P%,  •••,  Pn_iQ,  of  any  broken 
line  joining  a  point  P  of  space  with  a  second  point  Q  is  equal  to 
the  projection  on  L  of  the  directed  line-segment  PQ. 

Theorem  2  of  the  Introduction,  §  3,  may  be  extended  in  a 
similar  manner.  Let  the  student  state  and  prove  the  result. 

The  projection  of  a  point  P  on  a  plane  K  is  denned  as  the 
foot  of  the  perpendicular  dropped  from  P  on  K.  If  P  lies  in 
K,  it  is  its  own  projection  on  K. 

Let  a  plane  K  and  a  line  L  be  given.  If  L  is  not  perpen- 
dicular to  K,  the  projection  of  L  on  JfTis  denned  as  the  line  in 
which  the  plane  through  L  perpendicular  to  K  intersects  K. 
If  L  is  perpendicular  to  K,  the  projection  of  L  on  Kis  merely 
a  point,  the  point  in  which  it  meets  K. 

3.  The  Angle  between  Two  Directed  Lines.  Given  any  two 
indefinite  straight  lines  in  space  and  on  each  of  them  a  sense  ; 
to  define  the  angle  between  these  two  directed  lines. 

*In  drawing  this  figure,  we  have  placed  ourselves  in  space  so  that  the 
plane  through  L  and  P  appears  to  us  as  a  vertical  plane. 


PROJECTIONS.     COORDINATES  407 

If  the  lines  meet,  they  lie  in  a  plane.  The  angle  0  between 
them  shall  be  denned  as  the  angle  between  the  half-lines,  or 
rays,  issuing  from  their  point  of  intersection 
in  the  given  directions  (Fig.  2). 

If  the  lines  do  not  meet,  choose  an  arbitrary 
point  A  of  space,  and  draw  from  A  two  rays  F 

respectively  parallel  to  and  having  the  same 
senses  as  the  given  lines.     The  angle  between  the  given  lines 
shall  be  denned  as  the  angle  between  these  rays.* 

Remarks.  The  angle  0  is  the  angle  betiveen  the  directed  lines, 
not  the  angle  from  one  to  the  other.  It  has  always  a  positive 
or  zero  value,  i.e.  a  numerical,  and  not  an  algebraic,  value. 

It  is  futile,  in  the  geometry  of  space,  to  try  to  distinguish 
between  positive  and  negative  angles.  For  instance,  suppose 
that,  in  an  attempt  to  define  the  angle  from  one  of  two  directed 
lines  lying  in  a  plane  to  the  other,  we  should  agree  that  angles 
measured  in  the  counter-clockwise  sense  are  to  be  considered 
positive  and  those  measured  in  the  clockwise  sense,  negative. 
Then  the  angle/row  the  one  directed  line  to  the  other,  if  viewed 
from  a  certain  side  of  the  plane,  would  appear  positive ;  but, 
viewed  from  the  other  side  of  the  plane,  the  same  angle  would 
be  negative.  Viewing  the  angle  from  one  side  of  the  plane  is 
as  justifiable  as  viewing  it  from  the  other,  since  the  plane  is  im- 
mersed in  space  and  not  displayed  on  a  blackboard  or  on  the 
page  of  a  book.  Consequently,  we  should  still  be  at  a  loss  as 
to  whether  the  angle  is  positive  or  negative. 

There  are  two  angles  between  the  rays  shown  in  Fig.  2,  namely, 
0  and  360°  —  0.  One  of  these  is  necessarily  less  than  or  equal 
to  180°.  It  is  this  angle  which  we  agree  to  take  as  the  angle  be- 
tween the  directed  lines. 

*  For  example,  in  Fig.  1,  the  line  of  PQ,  directed  from  P  to  Q,  and  L, 
directed  to  the  right,  are  two  directed  lines.  The  angle  between  them  is 
the  angle  6  constructed  by  choosing  A  on  the  first  line,  at  P,  and  by  draw- 
ing through  P  the  line  L'  parallel  to  and  having  the  same  sense  as  L. 


408  ANALYTIC   GEOMETRY 

EXERCISES 

1.  Fasten  a  sheet  of  paper  to  the  floor  with  one  edge  against 
the  wall,  and  tack  a  second  sheet  to  the  wall  with  one  edge 
along  the  floor.     Draw  on  each  sheet  a  directed  line  so  that  the 
two  lines  meet.     (It  is  wise  to  draw  the  lines  before  fixing  the 
sheets  in  position.)     Crease  a  third  sheet  of  paper  so  as  to  form 
an  angle  which  will  just  fit  between  the  two  directed  rays.     By 
measuring  this  angle  with  a  protractor  determine  the  angle  be- 
tween the  two  directed  lines. 

2.  EepeatEx.  1  with  two  directed  lines  differing  widely  in 
position  from  the  first  two  chosen. 

3.  What  is  the  angle  between  the  two  lines  of  Ex.  1,  if  the 
sense  on  one  of  them  is  reversed  ? 

4.  By  the  method  of  Ex.  1  find  the  angle  between  two 
directed  lines,  one  011  the  floor  and  one  on  the  wall,  if  the  two 
lines  do  not  meet. 

5.  Prove  that,  if  L  and  L'  are  any  two  lines  in  space  and  any 
plane  F  is  passed  through  L,  there  will  be  a  plane  W  through 
L'  perpendicular  to  F.     That  is,  show  that  the  above  method  is 
applicable  to  the  problem  of  determining  the  angle  between 
any  two  directed  lines. 

4.   Value  of   the  Projection  of    a    Directed   Line-Segment. 

Assign  to  a  line  L  of  space  a  sense  and  adopt  a  unit  of 
length  for  all  measurements  in  space.  Then  a  directed 
line-segment  AB  on  L  is  represented  by  an  algebraic  num- 
ber, equal  numerically  to  the  length  of  AB  and  positive  or 
negative  according  as  the  direction  from  A  to  B  is  the  same 
as,  or  opposite  to,  the  direction  given  to  L]  cf.  Introduc- 
tion, §  2. 

In  particular,  to  the  projection  MN  on  L  of  a  directed  line- 
segment  PQ  corresponds  a  certain  algebraic  value  or  number, 
which  we  can,  without  confusion,  denote  also  by  MN.  Clearly, 

Proj.  PQ  =  -  Proj.  QP. 


PROJECTIONS.     COORDINATES  409 

Let  the  length  |  PQ  \  of  the  directed  line-segment  PQ  be  given 
and  also  the  angle  6  which  the  line  of  PQ,  directed  from  P  to 
Q,  makes  with  the  directed  line  L.  If 
PQ,  lies  in  a  plane  with  L,  we  know  from 
Plane  Trigonometry  that 

(1)     MN  =  Proj.j  PQ  =  |  PQ  \  cos  6. 

The  general  case,  in  which  PQ  is  not  in  FlQ  3 

a  plane  with  L,  is  shown  in  Fig.  1.     Con- 
sider the  projection,  PR,  of  PQ  on  the  line  L'  through  P, 
parallel  to  and  having  the  same  sense  as  L.     Since  PQ  lies  in 
a  plane  with  L',  we  have  the  previous  case.     Consequently, 

PR  =  \PQ\  cos6. 

But  PR  =  MN  and  thus  formula  (1)  is  established  in  the  gen- 
eral case. 

EXERCISES 

1.  Draw  Fig.  1  for  various  positions  of  P  and  Q  and  in  each 
case  verify  formula  (1). 

2.  By  application  of  (1)  verify  that  Proj.  PQ  =  -  Proj.  QP. 

3.  If  P  and  Q  lie  in  a  plane  perpendicular  to  L,  M  and  N 
coincide  and  MN=  0.     Prove  this  by  applying  formula  (1). 

4.  Prove  that  the  directed  line-segments  MN  and   M'N1, 
which  are  the  projections  on  L  of  two  directed  line-segments 
PQ  and  P'Q'  on  the  same  line,  are  proportional  to  PQ  and  PQ' : 

MN  =  PQ 
M'N'     P'Q'' 

5.  Prove  that  the  theorem  of  the  preceding  exercise  is  true 
if  PQ  and  P'Q'  are  on  parallel  lines. 

5.  Coordinates.  Three  directed  lines  drawn  through  a  point 
0  of  space,  so  that  each  is  perpendicular  to  the  other  two, 
form  a  system  of  rectangular  coordinate  axes.  The  coordinates 
of  a  point  P  of  space  with  respect  to  the  system  of  axes  are 


410  ANALYTIC   GEOMETRY 

defined    as   the   numbers   which   represent   algebraically   the 
projections   of   the   directed   line-segment   OP  on   the   three 
directed    lines.      Thus,    if    Ox,    Oy,    Oz 
denote  the  three  directed  lines  and  x,  y,  z 
stand  for  the  coordinates  of  P, 

x  =  Proj.o,  OP,         y  =  Pro].,*  OP, 


The   projections  of    OP  on   the  three 
directed  lines  can  be  constructed  by  pass- 
ing planes  through  P  perpendicular  to 
the  three  lines  (Fig.  4).     These  form  with 
the  planes  of  the  lines  a  rectangular  parallelepiped,  or  box. 
The  directed  edges  of  the  box  which  issue  from  0  are  the 
projections  of  OP. 

Every  point  P  of  space  has  unique  coordinates  (x,  y,  z).  Con- 
versely, if  any  three  numbers  x,  y,  z  are  given,  there  is  a 
unique  point  P  having  these  numbers  as  its  coordinates.  This 
point  can  be  located,  either  by  constructing  a  box  or,  more 
simply,  by  laying  off  OM  =  x  on  the  axis  of  x,  then  MN=  y 
on  a  parallel  through  M  to  the  axis  of  y,  and  finally  NP  =  z  on 
a  parallel  through  N  to  the  axis  of  z,  as  shown  in  Fig.  4.  It 
is  to  be  remembered  that  OM,  MN,  and  NP  are  directed  line- 
segments.  The  direction  of  OM,  for  example,  is  the  same 
as,  or  opposite  to,  that  of  Ox,  according  as  the  number  x 
is  positive  or  negative.  The  figure  is  drawn  for  the  case  that 
x,  y,  z  are  all  positive. 

The  point  O  is  the  origin  of  coordinates,  the  directed  lines 
Ox,  Oy,  Oz  are  the  coordinate  axes,  and  the  planes  xOy,  yOz, 
zOx  are  the  coordinate  planes.  The  origin  has  the  coordinates 
(0,  0,  0),  a  point  on  a  coordinate  axis  always  has  two  of  its 
coordinates  zero,  and  a  point  in  a  coordinate  plane  always  has 
one  zero  coordinate.  Thus  the  point  on  the  axis  of  y  three  units 
distant  from  0  in  the  positive  direction  has  the  coordinates 
(0,  3,  0),  and  the  point  in  the  (y,  z)-plane,  whose  coordinates 
in  that  plane  are  y  =  2,  z  =  3,  has  the  coordinates  (0,  2,  3). 


PROJECTIONS.     COORDINATES  411 

Octants.  Bounded  by  the  coordinate  planes  there  are  eight 
regions,  called  octants.  It  is  clear  that,  if  the  x-cob'rdinate  of 
one  point  of  an  octant  is,  for  example,  negative,  the  ^coordi- 
nates of  all  points  of  the  octant  are  negative ;  similarly,  for 
the  y-  and  z-cob'rdinates.  Thus  we  can  speak  of  the  octant 
(  — ,  +,  +),  and  mean  thereby  that  octant  in  which  the 
a^coordinate  of  every  point  is  negative  and  the  y-  and  z-coor- 
dinates  are  both  positive.  The  octant  (+,  +,  +)  is  known  as 
the  first  octant ;  we  make  no  attempt  to  number  the  others. 

Figures.  In  drawing  a  figure  in  a  plane  to  represent  a 
figure  in  space,  we  make  use  of  what  is  known  as  a  parallel 
projection.  The  axes  of  y  and  z  are  represented  by  two  per- 
pendicular lines  and  the  axis  of  a;  by  a  line  drawn  in  a  con- 
venient direction.  All  distances  in  the  (y,  z)-plane  or  in  any 
parallel  plane  are  drawn  to  scale,  so  that  a  figure  in  such 
a  plane  appears  as  it  actually  is  in  space.  Distances  on  or 
parallel  to  the  axis  of  x  are  foreshortened  a  convenient  amount. 

The  direction  of  the  line  representing  the  axis  of  x  and  the 
amount  of  foreshortening  along  this  axis  depend  largely  on 
the  figure  in  space  which  is  to  be  represented.  In  general, 
however,  we  shall  draw  the  line  representing  the  axis  of  x  at 
an  angle  of  120°  with  that  representing  the  axis  of  y  and  take 
as  the  unit  distance  on  the  axis  of  x  three-fourths  the  unit 
distance  on  the  other  axes. 

Right-Handed  and   Left-Handed   Coordinate  Systems.     The 
system  of  axes  in  Fig.  4  is  the  one  we  shall  employ.     Another 
system  in  common  use  is  shown  in  Fig.  5.    The 
essential  difference  between  the  two  is  this  : 
If,  from  a  point  on  the  negative  axis  of  x,  we 
view  the  (y,  z)-plane,  the  direction  of  rotation 
from  the  positive  y-axis  to  the  positive  z-axis      / 
is  that  of  a  right-handed  screw  in  case  of  the    /v 
first  system  and  that  of  a  left-handed  screw  in 
case  of  the  second.     Accordingly,  the  system  we  are  using  is 
called  a  right-handed  system ;  the  other,  a  left-handed  system. 


412  ANALYTIC   GEOMETRY 

» 

Any  other  rectangular  system  of  axes  is  essentially  the 
same  as  one  or  the  other  of  these ;  that  is,  it  is  either  right- 
handed  or  left-handed,  by  the  above  test  (cf.  Ex.  7). 

EXERCISES 

1.  Plot  the  following  points,  drawing  the  line  representing 
the  axis  of  x  at  an  angle  of  120°  with  that  representing  the 
axis  of  y,  and  taking  1  in.  as  the  unit  on  the  axes  of  y  and  z 
and  |  in.  as  the  unit  on  the  axis  of  x. 

(a)    (0,3,0);  (6)   (0,1,3);  (c)  (2,5,0); 

(d)   (4,0,0);  («)    (0,  -2,0);  (/)  (4,1,3); 

(3)   (5,  -  2,  4) ;  (A)  (3,  2,  -  5) ;  (t)  (-  2,  3,  1J) ; 

(j)  (!,-!,-  3);  (*)(-2,4,-3);  (Z)  (-  1,  -  1,  -2). 

2.  Determine  the  coordinates  of  the  point  P  in  Fig.  4  when 
the  units  on  the  axes  are  taken  as  in  Ex.  1. 

3.  The  same  for  the  point  marked  by  the  period  in  "  Fig.  4," 
if  this  point  is  ^  a  unit  above  the  (x,  2/)-plane. 

4.  What  are  the  coordinates  of  the  projections  of  each  of 
the  following  points  on  the  coordinate  axes  ?     On  the  coordi- 
nate planes  ? 

(a)     (3,5,2);          (6)     (-3,2,-!);          (c)     (x,  y,  z). 

5.  What  equation  is  satisfied  by  the  coordinates  of  those 
points  and  only  those  points  which  lie  in  the  (y,  z)-plane? 
In  the  (z,  #)-plane  ?     In  the  (x,  y)-plane  ? 

6.  What  two  equations  are  satisfied  by  the  coordinates  of 
those  points  and  only  those  points  which  lie  on  the  ic-axis  ? 
On  the  ?/-axis  ?     On  the  2-axis  ? 

7.  Through  a  point  0  draw  three  mutually  perpendicular 
lines,  which,  when  directed,  are  to  serve  respectively  as  the 
axes  of  x,  y,  and  z.     Show  that  there  are  eight  possible  com- 
binations of  the  directions  which  can  be  given  to  the  lines  and 
that,  of  the  eight  resulting  systems  of  axes,  four  are  right- 
handed  and  four,  left-handed. 


PROJECTIONS.     COORDINATES 


413 


6.  Projections  of  a  Directed  Line-Segment  on  the  Axes.  Given 
the  points  Pl  and  P2  with  the  coordinates  (xl}  yl}  zt)  and 
(x2,  y2,  z2).  To  determine  the  projections  of  the  directed  line- 
segment  PiP2  on  the  coordinate  axes,  project  the  broken  line 
PiOP2  on  each  of  the  axes  in  turn.  Since  always 

Proj.  AP2  =  Proj.  P,0  +  Proj.  OP2, 
it  follows  that 

Proj.  PiP2  =  Proj.  OP2  -  Proj.  OPX. 

But  the  projections  of  OP2  and  OP:  on  the  three  axes  are,  by 
definition,  the  coordinates  of  P2  and  P^  Consequently,  the 
projections  of  the  directed  line-segment  PiP2  on  the  three 
axes  are,  respectively, 

(i)  «2  — «i      y*  —  y\      «s  — «i- 

By  passing  planes  through  the  points  Pl  and  P2  perpen- 
dicular to  the  three  axes,  we  obtain  on  the  axes  the  actual 
projections,  -X^-X^,  Y^Y^  Z^Z*, 
of  the  directed  line-segment 
PiP2.*  The  planes  also -deter- 
mine a  rectangular  parallele- 
piped, or  box,  whose  three 
dimensions  are  equal  to  the 
numerical  values  of  the  three 
projections.  Accordingly,  the 
edges  of  the  box,  when  properly  x* 
directed,  are  precisely  equal  to 
the  projections.  In  particular, 
the  three  edges  emanating  from  P1?  i.e.  the  directed  line- 
segments  PiR,  PiS,  PiT,  are  equal  respectively  to  the  three 
projections  X^Xz,  YiY2,  Z^. 

EXERCISES 

1.  Plot  P1PZ  when  P1  is  the  point  (6)  of  Ex.  1,  §  5,  and  P2 
is  (c).  Determine  the  projections  from  the  figure  and  verify 
by  applying  formulas  (1). 

*  To  keep  the  figure  simple,  only  X\Xi  and  Y\Yi  are  shown. 


FIG.  6 


414  ANALYTIC   GEOMETRY 

2.  The  same  when 

i)   Pi  is  (c)  and  P2  is  (/)  ;  ii)   Px  is  (/)  and  P2  is  (g). 

3.  The  projections  of  PiP2  on  the  axes  are  2,  —  5,  3  and 
those  of  P2P3  are  —  3,  2, 1.     What  are  the  projections  of  P:P3  ? 
Justify  your  answer. 

4.  If  the  points  P1}  R,  S,  T  of  Fig.  6  have,  respectively, 
the  coordinates  (2,  1,  3),  (5,  1,  3),  (2,  4,  3),  (2,  1,  6),  what  are 
the  coordinates  of  P2  ? 

5.  If  the  projections  of  PxP2  on  the  axes  are  3,  —  5,  —  2 
and  P!  has  the  coordinates  (2,  1,  3),  what  are  the  coordinates 
ofP2? 

7.  Distance  between  Two  Points.  Let  the  two  points  be 
the  points  P1}  P2  of  §  6.  Then  the  segment  PjP2  is  a  diagonal 
of  the  box  in  Fig.  6.  It  is  a  simple  matter  to  shtiw  that  the 
square  of  the  length  of  a  diagonal  equals  the  sum  of  the 
squares  of  the  lengths  of  the  edges : 

PiP*  =  P^  +  P^  +  PiT*. 
Hence  D*  =  (x2  -  xtf  +  (y,  -  y^  +  (z2  -  ztf, 


and  D  =  V(x2  -  xtf  +  (yt  -  y,)2  +  (z2  -  ztf. 

Inasmuch  as  it  is  the  squares  of  the  quantities  (1),  §  6, 
which  appear  here  under  the  radical,  it  is  immaterial  that 
these  quantities  have  algebraic  values,  i.e.  may  in  some  cases 
be  positive  and  in  others,  negative ;  cf.  Ch.  I,  §  3. 

EXERCISES 

1.  Find  the  distances  between  the  following  pairs  of  points, 
expressing  the  results  correct  to  three  significant  figures. 

(a)     (5,  1,  4),  (4,  3,  2) ;  (6)     (2, -1,3),  (-1, 1, -3)  ; 

(c)      (2,  -1,8),  (-2,  -3,5);       (d)     (3,  6,  -2),  (5,  -1,  4)  ; 
(e)     (2,  -3,5),  (-1,4,5);          (/)     (1,  2,  4),  (1,  -  3,  4). 

2.  Find  the  distances  of  each  of  the  following  points  from 
the  origin : 

(a)     (4,2,8);         (6)     (3,  -  5,  -  2) ;         (c)     (x,  y,  z). 


PROJECTIONS.     COORDINATES  415 

3.  Find  the  distances  of  each  of  the  points  of  Ex.  2  from 
the  coordinate  axes. 

4.  Find  the  lengths  of  the  projections  on   the  coordinate 
planes  of  the  line-segment  joining  the  points  (2,  3,  5)  and 
(5,  6,  7).     Draw  a  figure  showing  the  projections,  or  line- 
segments  equal  to  them. 

5-  What  equation  is  satisfied  by  the  coordinates  of  those 
points  and  only  those  points  which  lie  on  the  unit  sphere,  — 
the  sphere  whose  center  is  at  the  origin  and  whose  radius  is 
one  unit  ? 

8.  Mid-Point  of  a  Line-Segment.  Let  Pl  :  (ajt,  yl}  z^  and 
P2  :  (x2,  y2,  z2)  be  the  extremities  of  the  line-segment  PiP2.  If 
P  :  (x,  y,  z)  is  the  mid-point  of  PiP2,  the  directed  line-seg- 
ments PiP  and  PP2  are  equal  and  have,  therefore,  equal  pro- 
jections on  the  coordinate  axes.  Thus  we  have,  by  (1),  §  6, 

x  —  xl  =  x2  —  x, 
and  similar  equations  in  the  y-  and  z-coordinates.     Hence, 

m          r_?L±^2        v  —  yi+J*        *  —  *L±*I. 
~~~'  ~~'  ~~~ 


This  result  can  be  stated  in  words  as  follows  :  Tfie  coordi- 
nates of  the  mid-point  of  a  line-segment  are,  respectively,  the 
averages  of  the  corresponding  coordinates  of  the  end-points  of 
the  segment. 

EXERCISES 

1.  Determine  the  coordinates  of  the  mid-point  of  each  of 
the  line-segments  given  by  the  pairs  of  points  in  Ex.  1,  §  7. 
Draw  figures  and  check  your  answers. 

2.  Show  that  the  sum  of  the  squares  of  the  diagonals  of  the 
quadrilateral  whose  successive  vertices  are  at  the  four  points 
(5,  0,  0),  (0,  6,  0),  (1,  2,  3),  (3,  -  2,  8)  is  double  the  sum  of 
the  squares  of  the  line-segments  joining  the  mid-points  of  the 
opposite  sides.     N.  B.     The  four  points  do  not  lie  in  a  plane. 


416  ANALYTIC   GEOMETRY 

3.  Show  that  the  line-segments  joining  the  mid-points  of 
the  opposite  sides  of  the  quadrilateral  of  Ex.  2  intersect  and 
bisect  each  other. 

9.   Division  of  a  Line-Segment  in  a  Given  Ratio.     Let  it  be 

required  to  find  the  coordinates  (x,  y,  z)  of  the  point  P  divid- 
ing the  line-segment  PxP2  in  the  given  ratio  m,i/m2.  Since 
the  directed  line-segments  PjP  and  PP2  are  to  be  in  the  ratio 
m1/m2,  this  must  also  be  the  ratio  of  their  projections  on  any 
one  of  the  axes  (§4,  Ex.  4).  Accordingly,  we  obtain  the 
equation 

x  —  Xi  _  nil 

x2  —  x     m2 

and  similar  equations  involving  the  y-  and  z-cob'rdinates. 

Solving  these  equations  respectively  for  x,  y,  and  z  gives,  as 
the  required  coordinates  of  the  point  P, 


External  Division,  It  is  sometimes  of  value  to  have  at 
hand  formulas  giving  the  coordinates  (x,  y,  z)  of  a  point  P 
which  lies  on  the  line  of  PI  and  P2,  but  exterior  to  the  seg- 
ment PiP2,  and  whose  distances  to  PI  and  P2  are  in  a  given 
ratio  mi/m2,  not  equal  to  unity.  Let  the  student  show  that 
in  this  case  it  is  the  directed  line-segments  PXP  and  P2P 
which  are  in  the  given  ratio  m1/m2,  and  that  the  type  of 
equation  now  obtained  is 


x  —  x2     m2 
so  that  the  required  coordinates  of  P  are 


„  _  _ 

x  —  —  —  ,     y  — 


The  point  P  is  said  to  divide  the  segment  PjP2  internally, 
in  the  first  case  ;  externally,  in  the  second.  The  numbers  mi 
and  ra2  entering  into  the  ratio  of  division  do  not  have  to  be  the 


PROJECTIONS.     COORDINATES  417 

actual  lengths  of  the  corresponding  line-segments,  but  rnay  be 
any  numbers  proportional  to  these  lengths.* 

EXERCISES 

1.  Find  the  coordinates  of  the  point  on  the   line-segment 
joining  (2,  —  3,  6)  with  (5,  4,  —  2),  which  is  twice  as  far  from 
the  first  point  as  from  the  second.  Ans.    (4,  |,  |-). 

2.  Find  the  point  on  the  line  through  the  two  points  of,Ex. 
1,  which  is  outside  the  line-segment  bounded  by  them  and  is 
twice  as  far  from  the  first  point  as  from  the  second. 

3.  Find  the  point  which  divides  internally  the  line-segment 
from  (2,  3,  4)  to  (o,  -  3,  0)  in  the  ratio  3  :  4. 

4.  The  preceding  exercise  for  external  division. 

EXERCISES  ON   CHAPTER  XVII  t 

1.  Show  that  the  points  (2,  4,  3),  (4,  1,  9),  (10,  -  1,  6)  are 
the  vertices  of  an  isosceles  right  triangle. 

2.  Prove  that  the  tetrahedron  with  vertices  at  the  points 
(0,  0,  0),  (0,  1,  1),  (1,  0,  1),  (1,  1,  0)  is  a  regular  tetrahedron. 

3.  Show  that  the  points  (0,  0,  V2),  (1,  1,  0),  (0,  0,  -  V2), 
(—1,  —  1,  0),  (2,  —  2,  0)  are  the  vertices  of  a  regular  pyramid 
with  a  square  base. 

4.  Given  the  points  A,  B,  C  with  coordinates  (2,  —  3,  5), 
(4,  2,  3),  (6,  7,.l).     By  proving  that  AB+BC=AC,  show  that 
the  three  points  lie  on  a  line. 

5.  Show  that  the  three  points  of  Ex.  4  lie  on  a  line  by 
proving   that   their   projections   on   each   of    two   coordinate 
planes  lie  on  a  line.     Justify  this  method  of  proof. 

6.  Determine  the  point  on  the  axis  of  y  which  is  equidis- 
tant from  the  two  points  (3,  —  2,  4),  (—  2,  6,  5). 


*  Thus,  in  the  case  of  internal  division,  if  PiP  =  100  cm.  and  PP2  = 
26  cm.,  mi  and  m2  might  be  properly  and  wisely  chosen  as  4  and  1. 

t  Exercises  1-6,  14-18  of  Ch.  XIX,  §  1,  and  Exercises  1-8,  18-22  of 
Ch.  XX,  §  1,  may  be  introduced  here,  if  it  seems  desirable. 


418  ANALYTIC   GEOMETRY 

7.  Determine  the  point  in  the  (y,  z)  -plane  which  is  equi- 
distant from  the  three  points  (3,  0,  2),  (2,  3,  0),  (1,  0,  0). 

8.  Two  vertices  of  a  regular  tetrahedron  are  at  the  points 
(0,  0,  2V2),  (0,  2,  0).     If  the  other  two  vertices  lie  in  the 
(x,  y)-plane,  find  their  coordinates. 

9.  A   regular   pyramid,  of  altitude  h,  has   a  square  base 
whose  vertices  lie  on  the  axes  of  x  and  y  and  whose  edges  are 
of  length  a.     What  are  the  coordinates  of  the  vertices  of  the 
pyramid  ? 

10.  If  P  is  the  mid-point  of  the  line-segment  PiP2,  and  P 
and  P2  have  the  coordinates  (3,  —  2,  5)  and  ( —  2,  4,  3)  respec- 
tively, what  are  the  coordinates  of  Pj  ? 

11.  If  P  divides  the  line-segment  P^a  internally  in  the 
ratio  2 : 3,  and   P!   and  P  have   respectively  the  coordinates 
(1,  4,   3)   and   (3,  2,    —  1),  determine  the  coordinates  of  P2. 

Ans.   (6,  - 1,  -  7). 

12.  Find  the  ratio  in  which  the  point  B  of  Ex.  4  divides  the 
segment  AC  of  that  exercise.  Ans.   1 : 1. 

13.  A  point  with  ^coordinate  6  lies  on  the  line  joining 
the   two   points    (2,— 3,  4),  (8,  0,  10).     Find   its   other   two 
coordinates. 

Suggestion.     Determine  the  ratio  in  which  the  point  divides 
the  line-segment  bounded  by  the  two  given  points. 

14.  Find    the    point  in  which    the    line   joining   the   two 
points  (2,  —3,  1),  (5,  4,  6)  meets  the  (z,  o;)-plane. 

Ans.  (3f  0,  3}). 

15.  If   the  length  of   the  line-segment  PiP2  is  D  and  the 
lengths  of  its  projections  on  the  coordinate  planes  are  A>  Az> 
A>  show  that 

2  Z)2  =  D?  +  A2  -f  A2- 

16.  Show   that   the    lines    joining    the   mid-points   of   the 
opposite  sides  of  any  quadrilateral  ABCD  intersect  and  bisect 
each  other.     N.B.    The  points  A,  B,  C,  D  do  not  necessarily  lie 
in  a  plane. 


PROJECTIONS.    COORDINATES  419 

17.  Show  that  the  sum  of  the  squares  of  the  diagonals  of  any 
quadrilateral  is  twice  the  sum  of  the  squares  of  the  line-seg- 
ments joining  the  mid-points  of  the  opposite  sides. 

18.  Prove  that  the  center  of  gravity   (intersection  of  the 
medians)  of  the  triangle  with  vertices  at  (x1}  yi}  z^),  (x2>  y2,  z2), 
(03,  y3,  z3)  has  the  coordinates 


19.  Prove  that  the  lines  joining  the  vertices  of  a  tetrahedron 
with  the  centers  of  gravity  of  the  opposite  faces  all  go  through 
a  point  P,  which  divides  each  of  them  in  the  ratio  3  :  1. 

20.  Prove  that  the  lines  joining  the  mid-points  of  opposite 
edges  of  a  tetrahedron  all  go  through  a  point,  which  bisects 
each  of  them.     Show  that  this   point   is    identical   with  the 
point  P  of  Ex.  19. 


CHAPTER   XVIII 


DIRECTION  COSINES.    DIRECTION  COMPONENTS 

1.  Direction  Cosines  of  a  Directed  Line.  Given  a  directed 
line  L  in  space ;  to  find  a  means  of  determining  or  fixing  its 
direction. 

The  directed  line  L  makes  definite  angles,  a,  ft,  y,  with  the 
positive  axes  of  x,  y,  z,  respectively.     If  L  does  not  go  through 
the   origin,   0,  draw   L'   through    0 
parallel  to  L  and  agreeing  with  it  in 
sense.     Then  K,  ft,  y  are  equal  respec- 
-•£/     tively  to  the  angles  which  L'  makes 
with  the  axes  (Ch.  XVII,  §  3).     The 
angles  a,  ft,  y  are  called  the  direction 
angles  of  the  directed  line  L. 

Direction  Cosines.  The  cosines  of 
the  angles  a,  ft,  y,  namely  cos  a,  cos  ft, 
cos  y,  are  known  as  the  direction 

cosines  of  L.  Since  a,  ft,  y  are,  by  definition  (Ch.  XVII,  §  3), 
angles  between  0°  and  180°  inclusive,  they  are  uniquely  deter- 
mined when  their  cosines  are  given,  and  conversely.  Accord- 
ingly, we  can  use  either  the  direction  angles  or  the  direction 
cosines  to  fix  the  direction  of  L.  We  choose  the  direction 
cosines. 

Evidently,  two  directed  lines  which  are  parallel  and  have 
the  same  sense  have  the  same  direction  angles  and  the  same 
direction  cosines. 

Exercise.  If  two  lines  are  parallel  but  have  opposite  senses, 
show  that  the  direction  angles  of  one  are  the  supplements  of 

420 


FIG.  1 


DIRECTION   COSINES  421 

the  direction  angles  of  the  other,  and  that  the  direction  cosines 
of  one  are  the  negatives  of  the  direction  cosines  of  the  other. 

Example  1.     What  are  the  direction  cosines  of  the  positive 
axis  of  y  ? 

Here,  a  =  90°,         0=0°,         y  =  90° ; 

and  cos  a  =  0,         cos  (3  =  1,         cos  y  =  0. 

Example  2.      Find   the   direction    cosines    of   the    line   bi- 
secting the  angle  between  the  negative  axis 
of  y  and  the  positive  axis  of  z,  and  directed 
upward. 

In  this  case, 

a  =  90°,         £  =  135°,         y  =  45° ; 
cos  a  =  0,       cos  /3  =  —  1 V2,       cos  y  =|  V2.  FIG.  2 

THEOREM  1.     The  sum  of  the  squares  of  the  direction  cosines 
of  a  directed  line  is  equal  to  unity : 

(1)  cos2 a  +  cos2  (B  +  cos2  y  =  1. 

To  prove  this  theorem,  take  a  point  P :  (x0,  yQ,  z0)  on  Le 
(Fig.  1)  so  that  the  direction  from  Oto  P  will  be  the  direction 
of  L',  and  consider  the  projections  of  the  directed  line-segment 
OP  on  the  axes.  These  are  equal,  on  the  one  hand,  to  the  coor- 
dinates x0,  yQ,  z0  of  P  (Ch.  XVII,  §  5),  and  on  the  other,  to  the 
quantities  OP  cos  a,  OP  cos /3,  OP  cosy  (Ch.  XVII,  §  4). 
Hence 

(2)  x0  =  OP  cos  a,        y0  =  OP  cos  /3,         z0  =  OP  cos  y ; 


(3)  ,  , 

Thus  cos2  a  +  cos2  ft  +  cos2  y  =  X"  +  y\ 

But 


and  the  theorem  is  proved. 

We  have  shown,  then,  that  every  directed  line  has  definite 
direction  cosines,  the  sum  of  whose  squares  is  unity.     The  con- 


422  ANALYTIC   GEOMETRY 

verse  is  also  true:  Any  three  numbers,  the  sum  of  whose  squares 
is  unity,  are  the  direction  cosines  of  some  directed  line. 

Preliminary  to  proving  this,  we  revert 
to  the  proof  of  Theorem  1  and  choose 
the  point  P  in  particular  as  the  point 
in  which  the  ray  issuing  from  0  in  the 
direction  of  L'  meets  the  unit  sphere 
(Fig.  3).  Then  OP  =1,  or 


and  (2)  becomes 

FIG.  3  •  ° 

That  is,  the  direction  cosines  of  a  ray 

issuing  from  0  are  equal  to  the  coordinates  of  the  point  in  which 
the  ray  pierces  the  unit  sphere. 

The  desired  proof  is  now  simple.  If  there  are  given  any 
three  numbers,  x0,  y0,  z0,  the  sum  of  whose  squares  is  unity, 
they  will  be  the  coordinates  of  some  point  P  of  the  unit  sphere, 
and  hence  they  will  also  be  the  direction  cosines  of  a  certain 
directed  line,  namely,  the  line  L'  passing  through  0  and  P  and 
directed  from  0  to  P,  q.  e.  d. 

Example  3.  The  three  numbers  f ,  ^,  —  f  are  the  direction 
cosines  of  some  directed  line,  for 


The   direction   angles   of   the   line  are,   respectively,  48°  11', 
70°  32',  131°  49'. 

Example  4.     A  directed  line  makes  angles  of  60°  and  45° 
with  the  axes  of  x  and  y,  respectively.     What  angle  does  it 
make  with  the  axis  of  z? 
Here, 

cos  a  =  cos  60°  =  \,         cos  ft  =  cos  45°  =  £  V2. 
Hence 

(I)2  +  (i  V2)2  +  cos2  y  =  1        and        cos  y  =  ±  £. 


DIRECTION   COSINES  423 

Thus  y  =  60°  or  120°.  There  are,  then,  two  directed  lines 
making  the  given  angles  with  the  x-  and  ?/-axes.  The  one 
makes  an  angle  of  60°  with  the  z-axis  ;  the  other,  an  angle  of 
120°. 

Direction  Cosines  of  the  Line  through  Two  Points.  Let  PiP2 
be  a  directed  line-segment  lying  on  the  directed  line  L  and 
having  the  same  sense  as  L.  If  XiX2,  Y^Y^  ^1^2  are  the 
projections  on  the  axes  of  PtP2  (Ch.  XVII,  Fig.  6),  we  have, 
by  Ch.  XVII,  §  4, 


where  D  is  the  length  of  the  segment  PiP2. 
Hence 


(5)  cos  a  =      t,         cos  ft  =  ,         cos  y  = 

The  content,  of  these  equations  can  be  stated  in  words,  as 
follows  : 

THEOREM  2.  If  a  directed  line  L  is  given  and  on  L  any 
directed  line-segment  PjP2  having  the  same  sense  as  L  is  chosen, 
the  direction  cosines  of  L  are  equal  to  the  projections  of  PiP2  on 
the  axes,  each  divided  by  the  length  of  PiP2. 

If  P!  and  P2  have  the  coordinates  (xl}  y^  z^)  and  (xz,  y2,  z2), 
formulas  (5)  become,  by  Ch.  XVII,  §§  6,  7, 

(6) 


,  , 

D  D  D 


where          D  =  V(z2  -  xtf  +  (y%  -  ytf  +  (z2  -  ztf. 

These  are  the  formulas  giving  the  direction  cosines  of  the  line 
passing  through  P1  :  (xl}  y1}  Zj)  and  P2  :  (x2,  y2,  z2),  and  directed 
from  Px  to  P2. 

EXERCISES 

1.  What  are  the  direction  cosines  of  a  line  parallel  to  the 
axis  of  z  and  having  the  same  sense?  Having  the  opposite 
sense? 


424  ANALYTIC   GEOMETRY 

2.  A  line  bisects  the  angle  between  the  positive  axes  of  y 
and  z  and  is  directed  upward.     What  are  its  direction  cosines  ? 

3.  What  are  the  direction  cosines  of  a  directed  line  which 
lies  in  the  (z,  #)-plane  and  makes  an  angle  of  30°  with  the 
positive  z-axis  ?     Two  answers. 

4.  A  line  in  the  (a;,  y)-plane  has  the  slope  V3.     What  are 
its  direction  angles  and  direction  cosines,  if  it  is  directed  for- 
wards ? 

5.  Construct  the  directed  lines  through  the  origin  having  the 
following  direction  cosines.     What  are  the  direction  angles  ? 

(a)   -  1,  0,  0;  (6)  |,  ±V3,  0;       (c)   -  |VS,  0,  i; 

(d)  0,  -iV2,£V2;    (e)   -f,  -4,0;  (/)  iV3,  -^VS^VS. 

6.  Find  the  direction  angles  and  the  direction  cosines  of  a 
line  if 

(a)     cosa  =  l;  (6)    .cos£  =  — 1,     cos.y=|V3; 

(c)     cos  a  =  i,     cos  /3  =  —  |V2. 

7.  Find  the  direction  angles  and  the  direction  cosines  of  a 
directed  line  if 

(a)     <*=120°,  0  =  60°;  (6)     a  =  135°,  y  =  120° ; 

(<0     ^  =  |'y  =  |5  (d)     a  =  45°,y8  =  y; 

(e)  «  =  /3  =  y;  (/)     «=y=180°-/?. 

8.  Find  the  direction  cosines  of  the  line  passing  through  the 
origin  and  each  of  the  following  points,  and  directed  from  the 
origin  to  the  point : 

(a)  (2,  3,  6) ;     (6)  (4,  -1,  8);     (c)  (3,  -4,  0);     (ef)  (5,  8,  -1). 

9.  Find  the  direction  cosines  of  the  lines  determined  by  the 
pairs  of  points  in  Ex.  1  of  Ch.  XVII,  §  7,  if  each  line  is  directed 
from  the  first  of  the  given  points  to  the  second. 

10.  A  line-segment  P\P<i  has  the  length  6  and  the  line  of  Pl 
and  P2>  directed  from  Pl  to  P2,  has  the  direction  cosines  —  f , 
^,  f.  If  the  coordinates  of  Pv  are  (—3,  2,  5),  what  are  those 
of  P2? 


DIRECTION   COSINES 


425 


N 


FIG.  4 


2.  Angle  between  Two  Directed  Lines.  Let  it  be  required  to 
find  -the  angle  0  between  the  two  directed  lines,  LI  and  7/2, 
whose  direction  angles  are  «1}  fa,  yi 
and  «2,  /32,  y2. 

We  can  assume  without  loss  of 
generality  that  LI  and  L2  pass 
through  the  origin.*  Take  any 
point  P  :  (x0,  yQ,  z0)  on  Ll}  so  that  the 
direction  from  0  to  P  is  that  of  LI, 
and  draw  the  broken  line  OMNP, 
whose  directed  segments  OM,  MN,  AT/ 
NP  are,  respectively,  the  coordi- 
nates x0,  y0,  z0  of  P.  The  projection 
of  this  broken  line  on  L2  equals  the  projection  of  OP  on  Z*2 

(1)      Pro  j  .LiOP=  Proj  .Lt  0  M  +  Pro  j  .^  MN  +  Proj  .  Jf  NP. 
By  Ch.  XVII,  §  4, 

Proj.£2  OP=  OP 
Similarly,          Proj.Za  OM  =  \OM\  cos 

where  J?"  is  a  point  on  L2,  such  that  O^fiT  has  the  direction  of 
L2.  If  the  directed  line-segment  OM  has  the  direction  of  the 
positive  axis  of  a?,  as  is  the  case  in  the  figure,  we  have 

|  OM\  =  OM,  %  KOM=  02, 

Proj.z   0  M  =  0  M  cos  a%. 
If  OM  has  the  direction  of  the  negative  axis  of  x, 

|  OM\  =  -  OM        and         £  KOM=  180°  -  «2  ; 
in  this  case,  then, 

Proj.£2  OM  =  -  OM  cos  (180°  -  a2)  =  OM  cos  a* 
Consequently,  in  either  case,  we  have,  since  OM=  x0, 
Proj.z   OM=  x0  cos  o-j. 

*  For,  if  they  did  not,  we  could  consider  equally  well  the  angle  between 
the  two  parallel  lines  through  the  origin  having  respectively  the  same 
senses  as  the  given  lines. 


and  therefore, 


426  ANALYTIC   GEOMETRY 

Similarly, 


j.£2  MN=  y0  cos  ft2,        Proj.Xj  NP  =  z0  cos  y2. 
Thus  (1)  becomes 

(2)  OP  cos  0  =  x0  cos  02  +  2/0  cos  /?2  +  z0  cos  y2. 
By  (2),  §1, 

i^o  =  OP  cos  «!,        ?/0  =  OP  cos  /Si,        z0  =  OP  cos  y^ 

Substituting  these  values  in  (2)  and  dividing  through  by  OP, 
we  obtain,  finally, 

(3)  COS  6  =  COS  ttj  COS  «2  +  COS  Pi  COS  /32  •+-  COS  yt  COS  y2. 

We  have,  then,  the  result  :  The  cosine  of  the  angle  between 
two  directed  lines  equals  the  sum  of  the  products  of  the  corre- 
sponding direction  cosines  of  the  lines. 

Example.  Find  the  angle  between  the  two  directed  lines 
whose  direction  cosines  are,  respectively,  -|,  —  ^,  —  |  and  ^,  ^,  ^. 

2.  3  +  (-1).  2  +  (-2).  6  8 

Here      cos0=  -  i-i  -  3.7       -  —  =  ~2l' 

whence  0  is  found  to  be  112°  24'.* 

Parallel  and  Perpendicular  Directed  Lines.  LI  and  L2  are 
perpendicular  if  and  only  if  6  =  90°  or  cos  0  =  0,  that  is,  if  and 
only  if 

(4)  COS  «!  COS  «2  +  COS  (3i  COS  /?2  +  COS  y!  COS  yjj  =  0. 

In  words  :  Two  directed  lines  are  perpendicular,  if  and  only  if 
the  sum  of  the  products  of  the  corresponding  direction  cosines  of 
the  lines  is  equal  to  zero.  Thus  the  directed  lines  which  have 
the  direction  cosines  f  ,  —  |,  f  and  ^,  \^,  T5¥  are  perpendicular, 

since 

2.  2-1-  14  +  2.  5 


3-15 


=  0. 


*  It  is  to  be  remembered  that  the  angle  between  two  directed  angles  is 
an  angle  between  0°  and  180°  inclusive  ;  cf .  Ch.  XVII,  §  3. 


DIRECTION   COMPONENTS       .  427 

We  repeat  here  the  results  concerning  parallelism  obtained 
in  §  1.  The  directed  lines  LI  and  L2  are  parallel  and  have 
the  same  sense  if  and  only  if  they  have  equal  direction  cosines  : 

(5)  cos  «!  =  cos  «2,         cos  /?!  =  cos  (32,         cos  yi  =  cos  y2. 

On  the  other  hand,  they  are  parallel,  but  with  opposite  senses, 
when  and  only  when  the  corresponding  direction  cosines  are 
negatives  of  each  other  : 

(6)  COS  «!  =  —  COS  «2,     COS  PI  =  —  COS  /?2,     COS  yt  =  —  COS  y2. 

EXERCISES 

In  each  of  the  following  exercises  find  the  angle  between 
two  directed  lines  with  the  given  direction  cosines. 


-  T9T, 


V21     V2I     V21      V14     V14         V14 
7.   Show  that  three  directed  lines  with  the  direction  cosines 

12._3__4  4123  3  412 

T3~>          13'          T¥>  T^J    T5>   T3">  T"3"J         T3"'   TJ' 

are  mutually  perpendicular. 

.  8.   Find  the  angle  subtended  at  the  point  (5,  2,  3)  by  the 
points  (2,  0,  -  3),  (-  9,  7,  5).  Ana.  79°  1'. 

9.  Determine  the  angles  of  the  triangle  with  vertices  at  the 
points  (1,  0,  0),  (0,  2,  0),  (0,  0,  3). 

3.  Direction  Components  of  an  Undirected  Line.  The  quan- 
tities ^,  —  7-,  7-  are  the  direction  cosines  of  some  line,  properly 
directed,  and  the  quantities  —  f,  f,  —  -f-  are  the  direction  co- 
sines of  this  line,  oppositely  directed. 


1. 

2       3 

6  .        6         2 

7  J      7'      7'     " 

f.      2. 

2                1 
3>      '~   $> 

-| 

3. 

t,  H 

21.841 
!>     15  '        9>     ¥>      9 

4. 

3               12 
T3~>          T^ 

_4 

2 

1          3 

2 

1 

1 

vii' 

V14"'    V14 

V6' 

V6 

V6 

fl 

4 

2          1 

2 

1 

3 

428  ANALYTIC   GEOMETRY 

Both  sets  of  direction  cosines  are  proportional  to  the  quan- 
tities 2,  —  3,  6.  Consequently,  these  quantities  pertain,  not  to 
the  line  directed  in  the  one  sense  or  the  other,  but  to  the  line 
bare  of  sense,  i.e.  to  the  undirected  line.  We  call  them  direc- 
tion components  of  the  undirected  line. 

It  is  clear  that  instead  of  2,  —  3,  6  we  might  have  taken 
equally  well  -  2, 3,  -6,  or  4,  -  6, 12,  or  200,  -  300,  600,  since 
the  two  sets  of  direction  cosines  are  proportional  to  the  quan- 
tities in  any  one  of  these  triples.  In  other  words,  the  direc- 
tion components  of  the  undirected  line  are  not  uniquely  deter- 
mined. There  are  infinitely  many  sets  of  direction  components ; 
if  one  set  is  2,  —  3,  6,  all  are  given  by  the  quantities  2p,  —  3p, 
6p,  where  p  is  an  arbitrary  number,  not  zero. 

Conversely,  if  we  have  given  the  set  of  direction  components, 
2,  —  3,  6,  of  the  undirected  line,  and  divide  each  by  the  square 
root  of  the  sum  of  the  squares  of  the  three,  i.e.  by 


we  obtain  the  direction  cosines,  -^,  —  f ,  ^,  of  the  line  directed 
in  one  sense.  Those  of  the  line  directed  in  the  opposite  sense 
are  the  negatives,  —  2,  3,  —  6,  of  the  given  direction  com- 
ponents, each  divided  by  the  above  square  root. 

The  General  Case.  Let  a  line  L  be  given  and  on  it  the  arbi- 
trary directed  line-segment  PiP2  whose  projections  on  the 
axes  are  X^,  Yi  F2,  Z&  (Ch.  XVII,  Fig.  6).  The  direction 
cosines  of  L,  when  directed  in  the  sense  of  PiP2,  are,  by  (5), 
§1. 


D  D  D   ' 

if,  however,  L  is  oppositely  directed,  in  the  sense  of  PZP\,  they 
are 

/o\  XzXi  ^2^1  -^2-^1 

D   '  /)    '          Z)   ' 

where,  in  each  case, 

(3)  D  = 


DIRECTION   COMPONENTS  429 

The  two  sets  of  direction  cosines  are  proportional  to  the 

quantities  * 

(4)  X,X2,         TiTi,         Z,Z2. 

These  quantities  pertain  merely  to  the  undirected  line  L.     We 
call  them  a  set  of  direction  components  of  L. 

Since  the  quantities  (4)  are  the  projections  of  PiP2  on  the 
axes,  this  definition  can  be  stated  as  follows. 

DEFINITION.  A  set  of  direction  components  of  an  UNDIRECTED 
line  L  are  the  projections  on  the  axes  of  a  directed  line-segment 
on  L. 

Instead  of  XiX.2,  Yi  F2,  Z^Z^  we  might  have  taken,  as  direc- 
tion components  of  L,  X2Xi}  Y2Yi,  ^2-^1?  i-e-  the  projections  of 
P2Pi  on  the  axes  ;  or  3  X^X^,  3  Y\Y^  3  Z^ZZ)  i.e.  the  projec- 
tions of  a  directed  line-segment  on  L  having  the  'same  sense  as 
PiPz  but  three  times  the  length. 

There  are,  then,  infinitely  many  sets  of  direction  components 
for  L.  Any  two  sets  are,  however,  proportional.  For,  two  arbi- 
trary sets  consist  of  the  projections  on  the  axes,  -X^-Xj,  Y]YZ, 
Z^Z^  and  XJXZ',  TV  Y2',  ZiZ2',  of  two  arbitrary  directed  line- 
segments,  PtP2  and  P/P2',  on  L.  But  the  projections  of  P/P2' 
and  PiP2  on  any  line  are  in  the  same  ratio  as  P/P2'  and  PiP2 
(Ch.  XVII,  §  4,  Ex.  4),  and,  therefore, 

V'"V'        P'P'  V'V        P'P'  7  '7'        P'P' 

-A-i  -A2    _  Jr  j  .T;)  2  i   JL  2    _  -T  \  f  2  ^1  -^2    _  -*  1  -^2 


or 

(5)    XiXz'  =  p  XiX2,       YI  Y2'  =  p  YI  Y2,        Zi'Z^  =  p  ZiZz, 

where  the  factor  of  proportionality,  p,  is  P1'P2'/P1P2,  q.  e.  d. 

Not  all  three  direction  components  can  be  zero.  For,  if  X^X^ 
YYz,  ZtZ?  were  all  zero,  then,  by  (3),  D  =  \  PtP2  1  would  be 
zero.  But  this  is  absurd,  since  P:  and  P2  are  distinct  points. 

We  summarize  our  results  in  the  form  of  a  theorem. 

*  The  factor  of  proportionality  is,  in  the  first  case,  1/D  ;  in  the  second, 
-l/D. 


430 


ANALYTIC   GEOMETRY 


THEOREM  1.  An  undirected  line  L  has  infinitely  many  sets  of 
direction  components;  if  I,  m,  n  is  one  set,  all  the  sets  are  given  by 
pi,  pm,  pn,  where  p  is  an  arbitrary  number,  not  zero;  moreover, 
I,  m,  n  are  not  all  zero.  Any  line  parallel  to  L  has  the  same  sets 
of  direction  components  as  L. 

The  last  statement  has  not  been  explicitly  proved.  We  leave 
the  proof  to  the  student ;  cf.  Ch.  XVII,  §  4,  Ex.  5. 

Example  1.  Find  the  direction  components  of  a  line  parallel 
to  the  axis  of  z. 

Take  any  directed  line-segment  PxP2  on  the  line.  Its  pro- 
jections on  the  axes  are  0,  0,  PiP2,  or,  if  a  is  the  number 
representing  PiP2,  they  are  0,  0,  a.  One  set  of  direction  com- 
ponents is,  then,  0,  0,  a  (a  =f=  0) ;  a  simpler  set,  and  the  one 
generally  used,  is  0,  0,  1. 

Example  2.  A  line  bisects  the  angle  between  the  positive 
axes  of  y  and  z.  What  are  its  direction  components  ? 

The  projection  on  the  a/-axis  of  any  directed 
line-segment  PiP2  on  the  given  line  is  zero, 
and  the  projections  on  the  axes  of  y  and  z  are 
equal.  If  the  number  representing  both  the 
latter  projections  is  a,  a  set  of  direction  com- 
ponents for  the  line  is  0,  a,  a.  A  simpler  set 
is  0,  1,  1. 

Geometrical  Representation   of  Direction  Components.     The 


FIG.  5 


directed  line-segments  XiXz,  Y1Y2, 
jections  of  P:P2  on  the  axes,  rep- 
resent geometrically  the  set  of 
direction  components  (4)  of  L. 
Instead  of  them  we  prefer  to  use 
the  equal  directed  line-segments 
P!^,  PI S,  P^T,  issuing  from  Pj 
(Fig.  6).  These  form  what  we 
shall  call  a  directed  trihedral-, 
PiR,  PiS,  PiT  are  its  directed 
edges,  and  Pb  its  vertex. 


which  are  the  pro- 


/ 

PZJ 

/^L 

_s 

^ 

-} 

K 

/ 

1 

I 

x,  / 

JC.Z-. 

FIG.  6 


DIRECTION   COMPONENTS 


431 


The  directed  trihedral  P^RST  represents  the  set  of  direc- 
tion components  (4).  Any  second  set,  consisting  of  the  pro- 
jections on  the  axes,  Xj'-XV,  ri'iy>  Zi'Z.2',  of  any  second 
directed  line-segment  P//Y  on  L  is  represented  by  the  directed 
trihedral  P^-R'S'  T'.  For  the  two  directed  trihedrals  we  have, 
from  (5), 

(6)        P^R'^pP.R,         P1'S'  =  PP1S,         P1'T'  =  pP1T. 

Because  of  this  relationship  we  call  them  similar.  That  is, 
two  directed  trihedrals  are  similar,  if  homologous  directed  edges 
are  proportional,  i.e.  if  the  directions  of  the  three  edges  of 
one  trihedral  are  all  the  same  as,  or  all  opposite  to,  the  direc- 
tions of  the  three  edges  of  the  other,  and  if  the  lengths  of 
homologous  edges  are  proportional. 

Since  any  two  sets  of  direction  components  of  L  are  in  the 
relation  (5),  the  directed  trihedrals  representing  them  are  in 
the  relation  (6)  and  are,  therefore,  similar.  Consequently, 
the  directed  trihedrals  representing  the  infinitely  many  sets 
of  direction  components  of  L  are  all  similar. 

Construction  of  a  Line  with  Given  Direction  Components. 
Let  it  be  required  to  construct  the  line  L  passing  through  a 
given  point  P1  in  space  and  having 
the  direction  components  4,  —  3,  2. 

Construct  a  directed  trihedral 
Pi-RST  with  Pl  as  vertex  and  with 
edges  P!#,  P:#,  P^T  defined,  both 
in  length  and  direction,  by  the 
numbers  4,  —  3,  2.  Complete  the 
box  determined  by  the  trihedral 
and  draw  the  diagonal  PtP2  issuing 
from  Pj.  'The  line  of  this  diagonal 
is  the  required  line  L.  For,  the 
projections  of  PiP2  on  the  axes  FIG.  7 

have  the  values  4,  —  3,  2. 

Incidentally,  we  have  shown  that  the  triple  4,  —  3,  2  is 
actually  a  set  of  direction  components  of  some  line,  L.  We 


432  ANALYTIC   GEOMETRY 

proceed  to  show,  further,  that  L  and  the  lines  parallel  to  it 
are  the  only  lines  having  this  triple  or,  more  generally,  the 
triples  4  p,  —  3  p,  2  p,  p  =£  0,  as  direction  components. 

Evidently  L  is  the  only  line  through  Pl  with  the  direction 
components  4,  —  3,  2.  For,  these  components  determine  the 
trihedral  at  P±  uniquely,  the  trihedral  determines  the  box 
uniquely,  and  the  box  the  line.  , 

If  we  took  8,  —  6,  4  instead  of  4,  —  3,  2  as  the  given  direc- 
tion components,  the  resulting  trihedral  would  have  edges  with 
the  same  directions,  but  twice  as  long,  as  the  edges  of  the 
original  trihedral,  i.e.  it  would  be  similar  to  the  original  trihe- 
dral. The  diagonal  of  the  new  box  which  issues  from  P1 
would  be  on  a  line  with  the  diagonal  Pi-P2  of  the  old  box 
and  so  the  same  line  L  would  be  determined.  Similarly, 
if  any  multiple,  4p,  —  3p,  2p,  of  4,  —3,  2,  where  p  is  any 
positive  or  negative  number,  were  taken  as  the  direction  com- 
ponents. 

Finally,  if  we  start  from  a  new  point  P/,  it  is  clear 
that  the  line  fj  through  it  with  the  given  direction  com- 
ponents will  be  parallel  to  L  or,  in  case  P/  lies  on  L,  the 
same  as  L. 

The  reasoning  here  is  perfectly  general,  applying  to  any 
triple  of  numbers,  I,  m,  n,  not  all  zero.  The  result  is  the  fol- 
lowing converse  of  Theorem  1. 

THEOREM  2.  If  I,  m,  n  are  any  three  numbers,  not  all  zero, 
the  triples  of  numbers  pi,  pm,  pn,  where  p  is  arbitrary  but  not 
zero,  are  sets  of  direction  components  of  some  undirected  line  L 
and  of  the  lines  parallel  to  L,  and  of  these  lines  only. 

Remark.  If  one  direction  component  is  zero,  the  corre- 
sponding edge  of  the  directed  trihedral  disappears,  and  the 
box  becomes  a  rectangle,  with  L  along  its  diagonal.  If  two 
direction  components  are  zero,  the  directed  trihedral  becomes 
a  directed  line  and  L  lies  along  this  line. 


DIRECTION   COMPONENTS  433 

EXERCISES 

In  each  of  the  following  exercises  find  all  the  sets  of  direc- 
tion components  of  the  given  line  and  then  choose  from  them 
a  simple  set. 

1.  A  line  parallel  to  the  axis  of  x. 

2.  The  line  bisecting  the  angle  between  the  positive  axis 
of  x  and  the  negative  axis  of  z. 

3.  A  line  in  the  (x,  t/)-plane  having  the  slope  2. 

Ans.   p,2P,  0;     1,2,0. 

4.  A  line  in  the  (y,  z)-plane  making  an  angle  of  60°  with 
the  y-axis.     Two  answers. 

5.  A  line  making  equal  angles  with  the  three  coordinate 
axes. 

6.  The  line  through  the  origin  and  the  point  (2,  1,  3). 

7.  The  line  through  the  points  (2,  3,  5),  (4,  7,  8). 

8.  What  can  you  say  of  the  position  of  a  line  if  one  of  its 
direction  components  is  zero  ?     If  two  are  zero  ? 

In  each  of  the  following  exercises  construct  the  line  through 
the  given  point  with  the  given  direction  components. 

Point  Direction  Components 

9.  Origin,  3,  5,  2. 

10.  Origin,  2,  -  3,  6. 

11.  (2,4,3),  p,p,2P(P=t=Q). 

12.  (5,  -4,6),  3,0,  -1. 

13.  (2,  5,  -  3),  0,  1,  0. 

4.  Formulas  for  the  Use  of  Direction  Components.  Direction 
Components  of  the  Line  through  Two  Points.  Let  the  two  points 
-Pi '  0»i>  2/i>  2i)>  A  :  (X2>  Vii  %z)  be  given.  Since  the  projections  of 
PjP2  on  the  axes  are  (Ch.  XVII,  §  6), 

(l)  »a-a;i,       y*-y\,       za-zi, 


434  ANALYTIC   GEOMETRY 

these  three  quantities  are  a  set  of  direction  components  of  the 
undirected  line  passing  through  Pl  and  P2. 

Relationship  between  Direction  Components  and  Direction 
Cosines.  We  saw  in  §  3  that  the  direction  components  of  an 
undirected  line  are  any  three  numbers,  not  all  zero,  propor- 
tional to  the  direction  cosines  of  the  line,  directed  in  one  sense 
or  the  other. 

Conversely,  starting  with  the  arbitrary  set  of  direction 
components,  XiX2,  Y^Y^,  ZiZ2,  of  an  undirected  line,  and 
dividing  each  component  by  the  square  root  of  the  sum 
of  the  squares  of  the  three,  i.e.  by  the  quantity  D  given  by 
(3),  §  3,  we  obtain  the  direction  cosines  (1),  §  3  of  the 
line,  directed  in  one  sense.  And  dividing  the  negatives 
of  the  direction  components  by  the  same  square  root,  we 
get  the  direction  cosines  (2),  §  3  of  the  line,  directed 
oppositely. 

We  state  this  result  as  a  theorem. 

THEOREM.  Ifl,m,n  are  a  set  of  direction  components  of  an 
undirected  line  L,  the  direction  cosines  of  L,  when  given  a  sense, 
are 

cos  a  =  — 


VZ2  -f  ra2  +  n2 


(2)  cos  (3  = 


±  m 


VZ2  +  m2  + 
±n 


COSy  =  -  ; 

VZ2  +  m2  +  n2 

where  either  all  the  upper  signs,  or  att  the  lower  signs,  are  to  be 
chosen  according  to  the  sense  which  has  been  given  to  L. 

If  we  had  used,  as  the  direction  components  of  L,  the  arbi- 
trary set  pi,  pm,  pn  (p  ^  0)  instead  of  the  particular  set  I,  m,  n, 
the  same  formulas  (2)  would  have  resulted.  For  if  in  (1)  Z, 
m,  n  are  replaced  by  pi,  pm,  pn,  p  comes  out  as  a  factor  from 
the  square  root  in  the  denominator  of  each  fraction  and  can- 


DIRECTION   COMPONENTS  435 

eels  the  p  in  the  numerator,  so   that   the   fractions   are  left 
unchanged.* 

Example  1.  A  directed  line  has  the  direction  cosines  $, 
_  ^?  _  |.  What  are  the  direction  components  of  the  line, 
undirected  ? 

Obviously,  2,  —  1,  —  2  are  one  set  of  direction  components, 
the  one  generally  used  ;  all  sets  are  given  by  2p,  —  p,  —  2p, 
where  p  ^  0. 

Example  2.  An  undirected  line  has  the  direction  compo- 
nents 4,  —  3,  12.  What  are  the  direction  cosines  of  the  line 
when  directed  ? 

The  sum  of  the  squares  of  the  given  direction  components 
is  169.  Hence  the  direction  cosines  of  the  directed  line  are 
either  ^,  -  ^,  if,  or  -  T\,  T3¥,  -  |f  ,  depending  on  the  sense 
in  which  the  line  is  directed. 

Angles  between  Two  Undirected  Lines.  Between  two  di- 
rected lines  there  is  but  one  angle  0  such  that  0  <  0  <  180°. 
Between  two  undirected  lines  LI  and  L»  there  are,  in  general, 
two  such  angles,  as  can  be  seen  readily  from  a  figure.  The 
two  angles  are  supplementary  and  have,  therefore,  cosines. 
which  are  negatives  of  each  other. 

If  the  direction  components  of  Z»t  and  Lz  are  llt  mt,  HI  and 
12,  w2,  n-i,  these  cosines  are  given  by  the  formula 

(3) 


VZr  +  mi2  +  M!*  VZ22  +  ma2  +  «22 

To  establish  this  formula,  write  down  by  use  of  (2)  the  direc- 
tion cosines  of  each  line,  directed  in  either  sense,  and  then 
apply  formula  (3),  §  2. 

It  follows  from  (3)  that  the  two  lines  L{  and  Lz  are  perpen- 
dicular when  and  only  when 

(4)  l^  +  ??i1m2  +  w1wi8'=  0. 


*  If  p  is  negative,  it  is  —  p  which  comes  out  as  a  factor  from  each 
square  root  and  hence  each  sign  ±  becomes  T. 


436  ANALYTIC   GEOMETRY 

The  lines  L^  and  L2  are  parallel  if  and  only  if  their  direction 
components  are  proportional,  i.e.  if  and  only  if 

(5)  k  =  pli,        wi2  =  pW!,         n2  =  Pn1, 

where  p   is  a  number   not   0.     This    follows    directly    from 
Theorems  1,  2,  §  3. 

EXERCISES 

1.  From   the   direction   cosines   of  the   directed   lines   of 
Exs.  1,  4,  6,  §  2,  find  the  direction  components  of  the  lines, 
undirected. 

2.  From  the  direction  components  of  the  undirected  lines 
of  Exs.  9-13,  §  3,  find  the  direction  cosines  of  the  lines,  directed 
first  in  one  sense  and  then  in  the  other. 

3.  A  line  has  the  direction  components  2,  8,  9.     What  are 
its  direction  cosines,  if  it  is  directed  upwards  ? 

4.  Find  the  direction  components  of  the  lines  joining  the 
origin  with  the  points  (c),  (/),  (t),  (I)  of  Ex.  1  of  Ch.  XVII, 
§5. 

5.  Find  the  direction  components  of  the  lines  determined 
•by  the  pairs  of  points  in  Ex.  1  of  Ch.  XVII,  §  7. 

In  each  of  the  following  cases  determine  the  angles  between 
two  lines  with  the  given  direction  components.  First  test  the 
lines  for  parallelism  or  perpendicularity. 

6.  3,  4,  -  1 ;     5,  -  2,  7.  7.  4,  -  2,  6  ;  -  6,  3,  -  9. 
8.   2,  -  1,  3  ;     2,  1,  - 1.  9.    -  3,  4,  2 ;  5,  8,  1. 

10.  Show  that  the  line  joining  the  origin  to  the  point 
(2,  1,  1)  is  perpendicular  to  the  line  determined  by  the 
points  (3,  5,  -  1),  (4,  3,  -  1). 

5.  Line  Perpendicular  to  Two  Given  Lines.  If  two  lines, 
intersecting  in  a  point  P,  are  given,  there  is  a  single  line 
through  P  perpendicular  to  each  of  them,  namely,  the  line 
through  P  perpendicular  to  the  plane  determined  by  them. 


DIRECTION    COMPONENTS  437 

More  generally,  let  any  two  non-parallel  lines,*  LI  and  L2, 
with  the  direction  components  li}  m^  nt  and  1%,  m»,  n^  be 
given.  Let  L  be  a  line  perpendicular  to  each  of  themf 
and  let  it  be  required  to  find  for  it  a  set  of  direction  com- 
ponents, Z,  ra,  n. 

We  begin  with  a  special  case.  Let  2,  3,  1  and  1,  4,  2  be  the 
direction  components  of  Li  and  L2,  respectively.  Since  L, 
with  the  direction  components  I,  m,  n,  is  perpendicular  to  L± 
and  also  to  L2,  we  have,  by  (4),  §  4, 

21  +.  3m  +  7i  =  0, 


From  these  £tuo  homogeneous  linear  equations  it  is  impossible 
to  determine  uniquely  the  three  unknowns  I,  m,  n.  But  this 
was  to  be  expected.  For,  there  is  not  a  unique  set  of  direction 
components,  I,  m,  n,  of  L,  but  infinitely  many  sets. 

In  general,  then,  there  will  be  a  set  for  which  n  =  1.  To 
determine  the  values  of  m  and  n  for  this  set,  we  must  solve 
simultaneously  the  equations 

r2x  2Z  +  3m  +  l  =  0, 

I  +  4  m  +  2  =  0. 

The  solutions  are  I  =  4,  m  =  —  f.  Consequently,  one  set  of 
direction  components  of  L  is  -|,  —  f  ,  1.  A  simpler  set  is  2, 
-3,5. 

In  the  general  case,  since  L  is  perpendicular  to  both  Ll  and 
Z/2,  it  follows  that 

=  0, 


I  ^j  } 

=  0. 


*  From  now  on  we  drop  the  qualifying  adjective  "undirected,"  and 
speak  merely  of  lines  and  directed  lines,  as  usual. 

t  There  are  infinitely  many  common  perpendiculars  to  LI  and  i2- 
They  are,  however,  all  parallel  to  one  another  and  hence  the  direction 
components  of  any  one  of  them  will  be  the  direction  components  of  all 
the  others. 


438 


ANALYTIC   GEOMETRY 


Here,  too,  we  try  to  find  a  set  of  direction  components  /,  m,  n 
of  L,  for  which  n  =  1.     Then  equations  (3)  become 


(4) 

I2l  +  mzm  =  — 

The  solutions  of  these  equations  are 

(5)  I: 


These  values  for  I  and  m,  together  with  w  =  1,  form  a  set  of 
direction  components  of  Z/.     A  simpler  set  is 


or,  in  determinant  form, 


(6) 


m2 


m2 


For  the  special  case  first  treated,  these  determinants  have 
the  values 


or  6  -  4  =  2,  1  -  4  =  -  3,  8  -  3  =  5.  But  2,  -  3,  5  were  the 
direction  components  found,  and  thus  the  work  in  the  special 
case  is  checked. 

Rule  of  Thumb.  To  obtain  the  determinants  in  (6)  easily, 
write  the  two  given  sets  of  direction  components  under  one 
another  : 


12    ra2    w2. 

The  first  determinant  in  (6)  is  formed  from  the  second  and 
third  columns  of  this  array  ;  the  second  is  formed 
from  the  third  and  first  columns  —  not  the  first 
and  third  —  and  the  third  from  the  first  and  second 
columns.  Thus  the  sets  of  numbers,  23,  31, 
1  2,  represent  the  columns  used  in  the  three  de- 
FIG.'  8  terminants.  The  first  set,  2  3,  is  all  that  need 


DIRECTION   COMPONENTS  439 

be  remembered.  For,  by  advancing  the  numbers  of  this  set  ac- 
cording to  the  cyclic  order  1  2  3  1  2  •  •  •,  this  set  2  3  becomes 
3  1,  i.e.  the  second  set ;  and  advancing  the  second  set  3  1 
cyclicly,  we  get  the  third  set,  1  2. 

Critique.     Not  all  the  determinants  (6)  are  zero,  for  if 

l\Wv  —  ltfn<\  =  fti-pii  —  mtfii  =  nil2  —  nzli  =  0, 
then  I1:l2  =  m1:m<i  =  nl:  n2 

or  ^  :  mi :  n^  =  12 :  m2  :  n2 

and  hence  the  lines  LI  and  L2  would  be  the  same  or  parallel, 
which  is  contrary  to  hypothesis. 

In  obtaining  the  solution  (5)  of  equations  (4)  we  assumed, 
tacitly,  that  ^m2  —  Z2w<i  =£  0 ;  there  was,  then,  a  set  of  com- 
ponents I,  m,  n,  for  which  n  =  1.  If  ^m2  —  Z2wi  =  0>  at  least 
one  of  the  two  remaining  determinants  cannot  be  zero.  If, 
for  example,  nj^  —  nJi  =£  0>  there  will  be  a  set  of  components 
I,  m,  n,  for  which  m  =  1,  and  we  can  find  this  set  by  putting 
m  =  1  in  (3)  and  solving  the  resulting  equations  for  I  and  n. 

EXERCISES 

In  each  of  the  following  exercises  determine  the  direction 
components  of  a  line  which  is  perpendicular  to  each  of  two  lines 
having  the  given  direction  components.  Actually  solve  the 
equations  and  then  check  the  result  by  the  rule  of  thumb. 


1. 


4. 


7.  The  two  directed  lines  LI  and  Z2  passing  through  the 
origin  and  having  respectively  the  direction  cosines  ^A/2, 
^V2,  0  and  —  ^V2,  £V2,  0  are  perpendicular  to  each  other. 
Find  the  direction  cosines  of  a  third  line  Ls  through  the  origin, 
perpendicular  to  both  Ll  and  L2,  and  so  directed  that  LI,  L2,  L3 


(3, 

4, 

2; 

2    •)    ' 

6,     -3; 

3     I1' 

2,     -1; 

I1' 

2, 

3. 

'  12, 

-4,     -1. 

1  Q 
1  3, 

—  1 

(,         2. 

!2' 

1, 

-1; 

5     !°' 

1,     0; 

6     ^     ' 

o, 

2; 

U 

2, 

3. 

5>  jo, 

0,     1. 

li 

o, 

_  i 

440  ANALYTIC  GEOMETRY 

form  a  directed  trihedral  (with  edges  of  indefinite  length) 
which  is  right-handed  by  the  test  of  Ch.  XVII,  §  5. 

Ans.  0,  0,  1. 

8.  The  above  problem,  if  the  direction  cosines  of  L^  and  L2 
are,  respectively,  f ,  f ,  —  f ,  and  -^,  f,  ^.  Ans.  fy,  —  fy,  fy. 

6.  Three  Lines  Parallel  to  a  Plane.  Given  three  lines,  with 
the  direction  components  J1?  m^  n1}  L,  m^,  n2,  and  £3,  m3,  ns. 
The  lines  will  be  parallel  to  a  plane  or  will  lie  in  a  plane, 
if  and  only  if  there  is  a  line,  with  the  direction  components 
I,  m,  n,  which  is  perpendicular  to  each  of  them,  i.e.  if  and 

only  if 

lil  +  mini  +  n^n  =  0, 

(1)  I2l  +  m2m  +  ntfi  =  0, 
I3l  +  m3m  +  nsn  =  0. 

But,  by  Ch.  XVI,  §  10,  these  three  homogeneous  linear  equa- 
tions have  a  solution  for  I,  m,  n,  other  than  the  obvious  solu- 
tion 0,  0,  0,  when  and  only  when  the  determinant  of  their 
coefficients  vanishes : 

?!    m,i    ni 

(2)  1%    m?    r?2  =  0. 

k    ^3    n3 

We  have,  therefore,  the  theorem :  * 

THEOREM.  Three  lines  are  parallel  to  a  plane  or  lie  in  a 
plane,  if  and  only  if  the  determinant  of  their  direction  components 
has  the  value  zero. 

EXERCISES 

1.  Show  that  three  lines  through  the  origin  with  the 
direction  components  2,  —1,  5,  3,  2,  —4,  7,  0,  6  lie  in  a 
plane. 

*  The  proof  covers  not  only  the  general  case,  when  the  given  lines 
have  but  one  common  perpendicular  direction  and  the  equations  (1)  a 
one-parameter  family  of  solutions,  but  also  the  special  case  in  which  the 
given  lines  are  parallel,  when  the  lines  have  infinitely  many  common 
perpendicular  directions  "and  the  equations  (1)  a  two-parameter  family 
of  solutions. 


DIRECTION   COMPONENTS  441 

In  each  of  the  following  exercises  show  that  three  lines  with 
the  given  direction  components  are  parallel  to  a  plane. 

2.  3,1,2;     5,  -4,3;     1,6,1. 

3.  -1,1,2;     2,  -1,1;     1,1,8. 

EXERCISES    ON    CHAPTER    XVIII 

1.  Show   that  the    triangle    with   vertices    at    the   points 
(1,  3,  -  5),  (3,  4,  -  7),  (2,  5,  -  3)  is  a  right  triangle. 

2.  Prove  that  the  points  (2,  - 1,  5),  (3,  4,  -  2),  (6,  2,  2), 
(5,  —3,  9)  are  the  vertices  of  a  parallelogram. 

3.  Show  that  (2,  3,  0),  (4,  5,  - 1),  (3,  7,  1),  (1,  5,  2)  are 
the  vertices  of  a  square. 

4.  Prove  that  the  three  points  A,  B,  C  with  the  coordinates 
(5,  -2,  3),  (2,  0,  2),  (11, -6,  5)  lie  on  a  line  by  showing  that 
the  line  AB  has  the  same  direction  as  the  line  AC. 

5.  Show  that  the  two  points  (4,  -2,  —6),  (-6,  3,  9)  lie 
on  a  line  with  the  origin. 

6.  Show  that  two  points  fa,  y^,  z^,  (x.2,  y^  z<i)  lie  on  a  line 
with   the   origin  when  and  only   when  their  coordinates  are 
proportional :     x1:yl:zl  =  x2:  y2 :  z2- 

7.  Show   that   the   four  points  A,  B,  C,  D,  with   the   co- 
ordinates (3,  4,  2),  (1,  6,  2),  (3,  5, 1),  (4,  5,  0),  lie  in  a  plane  by 
proving  that  the  sum  of  the  angles  which  BC  and  CD  subtend 
at  angle  A  equals  the  angle  which  BD  subtends  at  A. 

8.  Find  the  projection,  on  a  directed  line  having  -£,  —  f ,  f 
as  its  direction  cosines,  of  the  directed  line-segment  joining  the 
origin  to  the  point  (5,  2,  4).  Ans.  4. 

Suggestion.     Use  the  method  of  §  2  or  employ  formula  (2) 
of  Ch.  XVII,  §  4. 

9.  Show   that   the   projection,   on  a   directed   line   having 
cos  «,  cos  /?,  cos  y  as  its  direction  cosines,  of  the  directed  line- 
segment  joining  the  origin  to  the  point  (x,  y,  z)  is 

x  cos  a  +  y  cos  /3  +  z  cos  y. 


442  ANALYTIC   GEOMETRY 

10.  Find  the  projection,  on  a  directed  line  with  the  direc- 
tion cosines  f  ,  f  ,  —  ^,  of  the  directed  line-segment  joining  the 
point  (3,  —  2,  —5)  to  the  point  (8,  0,  —  2).  Ans.  5. 

11.  The  previous  problem,  in  the  general  case. 

12.  Two  lines,  LI  and  L2,  have  the  direction  components  1, 
1,  0  and  0,1,   —  1,   respectively.     Find   the   direction   com- 
ponents of  a  line  which  is  perpendicular  to  Li  and  makes  an 
angle  of  30°  with  L*  Ans.  1,  -  1,  2. 

13.  Prove  that  each  two  opposite  edges  of  the  tetrahedron, 
with  vertices  at  the  points  (0,  0,  0),  (1,  1,  0),  (0,  1,  —1), 
(1,  0,  —  1),  are  perpendicular. 

14.  A    tetrahedron    has    three    pairs    of    opposite    edges. 
Prove  that,  if  the  edges  of  each  of  two  pairs  are  perpendicular, 
the  edges  of  the  third  pair  are  also  perpendicular.     Choose  the 
coordinate  axes  skillfully. 

15.  Prove  the  identity 


=  (V  +  Mi2  +  v^)(X22  +  tf  +  v22)  - 


In  the  following  exercises  X1?  ^  V],  X2,  /^  "2  (and  A3, 
jn3,  v3)  denote  the  direction  cosines  of  directed  lines,  Lj,  L2 
(and  L3),  which  we  can  assume  go  through  the  origin.  In 
solving  the  exercises,  the  identity  of  Ex.  15  will  be  found 
useful. 

16.    If  6  is  the  angle  between  LI  and  L2,  show  that 
sin2  0  =  (/*iv2  —  ftevi)2  +  (vjX2  —  vaXi)2  -f-  (Xiju,2  —  X2/Ai)2- 


17.  Prove  that  if  L^  and  Z/2  are  perpendicular,  the  direction 
cosines  of  their  common  perpendicular  L3  are 

±  G"-!^  —  /A2"l)»  ("lX2  —  V2X!),  ±  (Xj/A2  —  Xo/Xi). 

18.  Show   that,  if  the  plus  signs  are   taken  in  the   above 
formulas,  L3  will  be  so  directed  that  the  lines  L1}  L>2,  L3  will 


DIRECTION   COMPONENTS  443 

form  a  directed  trihedral  which  is  right-handed  by  the  test  of 
Ch.  XVII,  §  5. 

19.  Prove  that,  if  Ll9  L2,  and  Ls  are  mutually  perpendicular, 
the  determinant,  |  A  /u,  v  \,  of  their  direction  cosines  has  the 
value  4- 1  or  —  1,  according  as  the  directed  trihedral  consist- 
ing of  Ll}  L2,  L3  is  right-handed  or  left-handed. 


CHAPTER   XIX 
THE   PLANE 

1.   Surfaces  and  Equations.     Example  1.    The  equation 

x  =  5 

is  satisfied  by  the  coordinates  of  those  points  and  only  those 
points  which  lie  in  the  plane  parallel  to  the  (y,  z)-plane  and 
5  units  in  front  of  it.  We  say  that  the  equation  represents 
this  plane. 

Example  2.     Consider  the  equation 

x  =  y. 

The  points  in  the  (x,  y)-plane  whose  coordinates  satisfy  it  are 
the  points  of  the  line  L  bisecting  the  angle  between  the  posi- 
tive x-  and  y-axes.  Since  z  is  unrestricted  by  the  equation, 
the  points  in  space  whose  coordinates  satisfy  it  are  the  points 
which  lie  directly  above  or  below  L,  or  are  on  L,  i.e.  the 
points  of  the  vertical  plane  through  L.  The  equation,  then, 
represents  this  plane. 

Example  3.     The  equation 


represents,  in  the  (z,  x)-plane,  a  circle,  C,  with  its  center  at  the 
origin  and  of  radius  5.  But  the  equation  does  not  restrict  in 
any  way  the  value  of  y.  Consequently,  it  represents  in  space 
the  circular  cylinder  formed  by  drawing  through  each  point 
of  the  circle  C  a  line  parallel  to  the  axis  of  y  and  extending 
indefinitely  in  both  directions. 

Surfaces.     The  planes  and  the  cylinder  represented  by  the 
three  equations  considered  are  known  as  surfaces  ;  the  cylinder 

444 


THE    PLANE  445 

is  a  curved  surface  and  the  planes  are  plane  surfaces.     In  gen- 
eralizing the  foregoing  discussions  we  should,  then,  say  : 

An  equation  in  x,  y,  z  represents,  usually,*  a  surface.  The 
surface  consists  of  all  those  points  and  only  those  points  whose 
coordinates,  when  substituted  for  x,  y,  z  in  the  equation, 
satisfy  it. 

Shifting  the  point  of  view,  we  assume  now  that  it  is  a  sur- 
face, and  not  an  equation,  which  is  given.  Then  we  should 
say: 

The  equation  of  a  given  surface  is  an  equation  in  x,  y,  z  which 
is  satisfied  by  the  coordinates  of  every  point  of  the  surface  and  by 
the  coordinates  of  no  other  point. 

Problem  1.  Find  the  equation  of  the  sphere  whose  center 
is  at  the  origin  and  whose  radius  is  a. 

A  point  (x,  y,  z)  lies  on  this  sphere  if  and  only  if  the  square 
of  its  distance  from  the  origin  is  equal  to  a2  : 


z  =  a. 
Therefore,  this  is  the  required  equation. 

Problem  2.     Find  the  equation  of  the  plane  which  passes 
through  the  axis  of  x  and  makes  an  angle  of 
30°  with  the  (x,  y)-plane,  as  shown  in  Fig.  1. 

This  plane  intersects  the  (y,  z)-plane  in  the 
line  whose  equation  in  the  (y,  z)-plane  is 

z  =  tan  30°  y        or        a  =  W3y. 

FIG.  l 

But  this  equation,  considered  as  an  equation  in 

x,  y,  z,  leaves  x  unrestricted  ;   consequently,  it  represents  in 

space   the   given  plane,  i.e.  it  is  the  equation  of  the   given 

plane. 

*  An  equation  in  x,  y,  z  does  not  always  represent  a  surface.  For 
example,  the  equation  x2  -f  yz  =  0  represents  a  line,  namely,  the  z-axis  ; 
the  equation  x2  +  y2  -f  z2  =  0  represents  just  one  point,  the  origin  ;  and 
the  equation  x2  +  y2  +  z2  +  1  =  0  represents  noypoint  whatsoever. 


446  ANALYTIC   GEOMETRY 

EXERCISES 

What  does  each  of  the  following  equations  represent  ?    Draw 
a  figure  in  each  case. 

1.  2  =  0.  5.  2t/  +  3z  =  6.  9.  x2  +  2y2  =  4. 

2.  y  +  3  =  0.  6.  y  =  \x  +  b.  10.  t/2  — z2  =  9. 

3.  ic  +  y  =  0.  7.  a-2  +  yz  =  a2.  11.  s2  —  9z2  =  —  9. 

4.  2  —  2a  =  0.  8.  z2  =  2z.  12.  a2  +  yz  +  z2  =  4. 

13.  Which  of  the  surfaces  represented  by  the  above  equa- 
tions pass  through  the  origin?     Which  contain  a  coordinate 
axis? 

Find  the  equations  of  the  following  surfaces. 

14.  The  (y,  z)-plane. 

15.  The  plane  parallel  to  the  (a;,  y)-plane  and  3  units  above 
it. 

16.  The  plane  parallel  to  the  (z,  x)-plane  and  2  units  to  the 
left  of  it. 

17.  The  plane  bisecting  the  angle  between  the  (x,  y)-  and 
(y,  z)-planes  and  passing  through  the  first  octant. 

18.  The  plane  perpendicular  to  the  (x,  y)-plane  whose  trace  * 
on  that  plane  has  the  slope  3  and  the  intercept  2  on  the  axis 
of  y. 

19.  The  circular  cylinder  whose  radius  is  3  and  whose  axis 
is  parallel  to  the  cc-axis  and  passes  through  the  point  (0,  1,  2). 

20.  The  parabolic  cylinder  whose  rulings  are  parallel  to  the 
y-axis  and  whose  trace  on  the  (z,  #)-plane  is  a  parabola  with 
its  vertex  at  the  origin  and  its  focus  at  the  point  (2,  0,  0). 

21.  The  elliptic  cylinder  whose  rulings  are  parallel  to  the 
z-axis  and  whose  trace  on  the  (x,  y)-plane  is  an  ellipse  which 
has  its  center  at  the  origin,  its  foci  on  the  .r-axis,  and  axes  of 
lengths  6  and  4. 

*  The  trace  of  a  surface  on  a  plane  is  the  line,  or  curve,  of  intersec- 
tion of  the  surface  with  the  plane. 


THE   PLANE  447 

22.  The  hyperbolic  cylinder  whose  rulings  are  parallel  to 
the  z-axis  and  whose  trace  on  the  (a;,  i/)-plane  is  a  rectangular 
hyperbola  with  its  center  at  the  origin  and  foci  at  the  points 
(0,  ±2,0). 

23.  The  sphere  whose  center  is  at  the  point  (1,  0,  0)  and 
whose  radius  is  unity. 

24.  The  sphere  whose  center  is  at  the  point  (1,  2,  —  3)  and 
whose  radius  is  5. 

2.  Plane  through  a  Point  with  Given  Direction  of  its  Nor- 
mals. Let  there  be  given  a  point  P,0  with  the  coordinates 
(XQ,  y0,  z0)  and  a  line  L  with  the  direction 
components  I,  m,  n.  Through  PQ  perpen- 
dicular to  L  there  is  just  one  plane.*  We 
propose  to  find  its  equation. 

Let  P :  (#,  y,  z)  be  any  point  of  the  plane 
other  than  P0.  Then  it  determines  with  P0 
a  line,  PoP,  which  is  perpendicular  to  the  FIG.  2 

line  L.     Since,  by  Ch.  XVIII,  §  4, 

a  —  z0,      'y-y0,       Z  —  ZQ 

are  the  direction  components  of  P0P  and  I,  m,  n  are  the  direc- 
tion components  of  L,  it  follows,  by  (4),  Ch.  XVIII,  §4,  that 

(1)  I(x-x0)  +  m(y-y0)  +  n(z-z0)=0. 

Conversely,  if  P:  (x,  y,  z)  be  any  point  other  than  P0  whose 
coordinates  satisfy  equation  (1),  this  equation  says  that  the  line 
PQP  is  perpendicular  to  L  and  hence  that  P  lies  in  the  plane. 

The  coordinates  x0,  y0,  z0  of  the  excepted  point,  P0,  obviously 
satisfy  equation  (1).  We  have  shown,  then,  that  this  equa- 
tion is  satisfied  by  the  coordinates  of  those  points  and  only 
those  points  which  lie  in  the  plane.  Hence  it  is  the  equation 
of  the  plane. 

There  are  infinitely  many  lines  L  perpendicular  or,  as  we 
say,  normal  to  the  plane,  and  they  are  all  parallel  to  one  an- 

*  The  figure  is  drawn  for  the  special  case  in  which  L  passes  through  PQ. 


448  ANALYTIC   GEOMETRY 

other.  It  is  their  common  direction  or,  analytically,  their 
common  direction  components,  which  are  essential.  Accord- 
ingly, we  speak  of  (1)  as  the  equation  of  a  plane  through  a  given 
point  with  a  given  direction  of  its  normals. 

EXERCISES 

In  each  of  the  following  exercises  find  the  equation  of  the 
plane  through  the  given  point  with  the  given  direction  of  its 
normals. 

Point  Direction  Point  Direction 

1.  (2,1,3),  1,1,  -2.  5.   Origin,  3,  -2,0. 

2.  (-5,  3,  4),      -  2,  2,  1.  6.    (5,  -  8,  2),     0,  1,  0. 

3.  (4,  -3,2),      5,0,3.  7.    (3,1,0),         0,0,1. 

4.  Origin,  2,  —  3,  5.  8.    Origin,  I,  m,  n. 

9.  Find  the  equations  of  the  three  planes  which  pass 
through  the  point  (5,  6,  —  3)  and  are  parallel  respectively  to 
the  coordinate  planes. 

10.  How  is  a  plane  situated  if  one  of  the  direction  com- 
ponents of  its  normals  is  zero  ?  If  two  are  zero  ? 

3.  The  General  Equation  of  the  First  Degree.  Since  any 
plane  can  be  determined  by  one  of  its  points  and  the  direction 
components  of  a  normal,  the  result  of  the  preceding  paragraph 
embraces  all  planes.  Moreover,  equation  (1)  of  that  paragraph 
is  of  the  first  degree  in  x,  y,  z.  We  have  thus  proved  the 
theorem:  Every  plane  can  be  represented  analytically  by  a 
linear  equation  in  x,  y,  z. 

Given,  conversely,  the  general  equation  of  the  first  degree  in 
x,  y,  z,  namely, 
(1)  Ax  +  By  +  Cz  +  D  =  0, 

where  A,  B,  C,  D  are  any  four  constants,  of  which  A,  B,  G 
are  not  all  zero.* 

*  In  dealing  with  equation  (1),  here  and  henceforth,  we  shall  always 
assume  that  J.,  J5,  C  are  not  all  zero. 


THE   PLANE  449 

Let  (XQ,  y0)  z0)  be  a  point  whose  coordinates  satisfy  equation1 

(1)  = 

(2)  Ax0  +  By0+Cz0  +  D  =  0. 

Subtracting  equation  (2)  from  equation  (1)  we  obtain 

A(x  -  x0)  +  B(y  -  y0~)  +  C(z  -  20)  =  0. 
But  this  equation  is  of  the  form  (1),  §  2,  where 
I  :  m  :  n  =  A  :  B  :  C. 

Therefore  it  represents  a  plane  which  has  A,  B,  C  as  the 
direction  components  of  its  normals.  This  result  we  state  as  a 
theorem. 

THEOREM.  The  general  linear  equation  (1)  always  represents 
a  plane.  The  coefficients  A,  B,  G  are  the  direction  components 
of  the  normals  to  the  plane. 

Example  1.     The  equation 


represents  a  plane  whose  normals  have  2,  —  3,  4  as  direction 
components.    The  point  (2,  2,  2),  for  example, 
lies  in  the  plane,  since  when  we  set  x  =  2, 
y  =  2  in  the  equation,  we  find  z  =  2  as  the 
value  of  z. 

We  obtain  a  rough  plot  of  the  plane  by 
constructing  the  point  (2,  2,  2)  and  the  line 
through  it  with  the  direction  components  2,  '* 

—  3,  4,  and  by  drawing  then,  as  accurately 
as   possible,  the   plane   through   the   point  perpendicular   to 
the  line. 

Example  2.     The  equation 

x  =  2,         or         a  +  0y  +  0«=2, 

is  the  equation  of  a  plane  having  the  lines  with  the  direction 
components  1,  0,  0  as  normals.  But  these  lines  are  parallel 
to  the  axis  of  x  and  hence  the  plane  is  parallel  to  the  (y,  z)- 
plane.  In  particular,  it  is  two  units  in  front  of  that  plane. 


450  ANALYTIC   GEOMETRY 

1  Remark.  In  the  proof  of  the  theorem  it  was  assumed  that 
there  is  always  a  point  whose  coordinates  satisfy  equation  (1). 
This  assumption  is  easily  justified.  By  hypothesis,  at  least 
one  of  the  three  coefficients  A,  B,  C  is  not  zero.  Suppose  that 
C  ^  0.  Then  equation  (1)  can  be  solved  for  z : 

Ax  +  By  +  D 
~C~ 

Giving  to  x  and  y  definite  values,  XQ  and  y0,  we  obtain  for  z  from 
this  equation  a  definite  value,  z0.  Then  the  point  (XQ)  y0,  z0) 
has  coordinates  which  satisfy  (1).  For  example,  if  XQ  =  0  and 
yQ  =  0,  then  z0  =  —  D/C,  and  the  point  is  (0,  0,  —  D/C). 

EXERCISES 

In  each  of  the  following  exercises  determine  the  direction 
components  of  the  normals  to  the  given  plane  and  the  coordi- 
nates of  a  point  lying  in  it.  Construct  the  plane  by  the  method 
of  Example  1. 

1.  3x  +  5y  +  Qz-5  =  Q.'   5.  2x  +  3y  —  5  =  0. 

2.  2x  —  y  +  2  —  3  =  0.      6.  3x  —  2  z  -  4  =  0. 

3.  4:X  +  2y  —  3z+6  =  0.    7.  5y  +  8  =  0. 

4.  5x  —  2y—  32  +  4  =  0.    8.  2  z  -  7  =  0. 

4.  Intercepts.  Let  a  plane  be  given  by  means  of  its  equation. 
A  simple  method  of  plotting  the  plane,  in  case  it  cuts  the  axes 
in  three  distinct  points  (one  on  each  axis),  consists  in  determin- 
ing from  the  equation  the  coordinates  of  these  three  points 
and  then  in  plotting  the  points  and  constructing  the  plane 
through  them. 

The  point  of  intersection  of  a  plane,  for  example, 
(1)  2z-32/+4z-6  =  0, 

with  the  axis  of  x  has  its  y-  and  z-coordinates  both  equal  to 
zero.     Consequently,  to  find  the  ^-coordinate  of  the  point,  we 
have  merely  to  set  y  =  0  and  z  =  0  in  the  equation  of  the  plane 
and  to  solve  for  x.     Thus,  in  this  case,  we  have 
2  a -6  =  0,         or        a  =  3. 


THE    PLANE  451 

The  point  of  intersection  of  the  plane  (1)  with  the  axis  of  x 
is,  then,  (3,  0,  0).  In  a  similar  manner  we  find  (0,  —  2,  0)  and 
(0,  0,  |)  as  the  points  of  intersection  of  the  plane  with  the  axes 
of  y  and  z  respectively.  By  plotting  these 
three  points  and  joining  them  by  lines,  we 
obtain  a  good  representation  of  the  plane. 

The  numbers  3,  —  2,  f  are  known  as  the 
intercepts  of  the  plane  (1)  on  the  axes  of  x,  y,  z, 
respectively.  That  is,  the  intercept  of  a  plane 
on  the  axis  of  x  is  the  x-coordinate  of  the 
point  in  which  the  plane  meets  the  axis  of  x. 
The  intercepts  on  the  axes  of  y  and  z  are  FIG.  4 

similarly  defined. 

A  plane  which  passes  through  an  axis  or  is  parallel  to  an  axis 
has  no  intercept  on  that  axis.  Every  other  plane  has  definite 
intercepts  on  all  three  axes  and  these  intercepts  determine  the 
position  of  the  plane  unless  they  are  all  zero,  that  is,  unless  the 
plane  goes  through  the  origin. 

EXERCISES 

Determine  the  intercepts  of  the  following  planes  on  the 
coordinate  axes,  so  far  as  they  exist,  and  construct  the 
planes. 

1.  2x+3y  +  4:Z  —  12  =  0.  6.  x-\-3y  —  z  =  0. 

2.  3x-2y  +  z-6  =  0.  7.  2  a  -  3  y  +  12  =  0. 

3.  x  +  y  —  z  —  2  =  0.  8.  3  y  +  4  z  —  6  =  0. 

4.  2x  +  oy  —  3  2  +  8  =  0.  9.  5  .T  +  2  z  =  0. 

5.  x  +  2y  +  z  +  3  =  0.  10.  3 a  +  5  =  0. 

5.   Intercept  Form  of  the  Equation  of  a  Plane.     Given  a 

plane  whose  position  is  determined  by  its  intercepts.  Let 
these  intercepts,  on  the  axes  of  x,  y,  z,  be  respectively  a,  6,  c. 
To  find  the  equation  of  the  plane  in  terms  of  a,  &,  c. 

We  have  the  problem  of  finding  the  equation  of  a  plane 


452  ANALYTIC   GEOMETRY 

through  the  three  points  (a,  0,  0),  (0,  6,  0),  (0,  0,  c).  Let  this 
equation  be 

(1)  Ax  +  By  +  Cz  +  D  =  0, 

where  the  values  of  A,  B,  C,  D  are  to  be  determined.  Since 
the  plane  does  not  go  through  the  origin,  D  ^  0.  Since  it 
contains  each  of  the  given  points,  the  following  equations  must 
hold: 

Aa  +  D  =  0,        Bb  +  D  =  0,         Cc  +  D  =  0. 

Hence  A  =  -  D/a,  B  =  -  D/b,  C=  —  D/c.  Substituting 
these  values  for  A,  B,  C  in  (1)  and  dividing  through  by  D,  we 
obtain 

(2)  *  +  |  +  ?  =  l. 
a     o      c 

That  this  is  the  desired  equation  can  easily  be  checked  by 
substituting  successively  in  it  the  coordinates  of  the  three 
points  in  question. 

Only  planes  which  intersect  the  axes  in  three  points  that  are 
distinct  can  have  their  equations  written  in  the  form  (2).  A 
plane  through  the  origin  is  an  exception,  because  at  least  one  of 
its  intercepts  is  zero  and  division  by  zero  is  impossible.  A 
plane  parallel  to  an  axis  is  also  an  exception,  since  it  has  no  in- 
tercept on  that  axis. 

EXERCISES 

Find  the  equations  of  the  planes  with  the  following  inter- 
cepts. 

1.  2,  3,  4.  3.    -  2,  4,  5.  5.     -  4,  -  6,  -  2. 

2.  2,  -3,  -1.          4.    -5,  -3,  2.  6.   2,  -8,-6. 

Find  the  equations  of  the  following  planes. 

7.  "With  intercepts  on  the  x-  and  y-axes  equal  to  2  and  3  and 
parallel  to  the  axis  of  z. 

8.  With   intercept  —  3  on  the  z-axis  and   parallel  to   the 
(x,  2/)-plane. 


THE   PLANE 


453 


9.  A  regular  quadrangular  pyramid  has  its  vertices  at  the 
points  (0,  0,  6),  (2,  0,  0),  (0,  2,  0),  (-  2,  0,  0),  (0,  -  2,  0). 
Find  the  equation  of  its  faces. 

10.  The  same,  if  the  vertices  are  at  the  points  (0,  0,  c), 
(a,  0,  0),  (0,  a,  0),  (-  a,  0,  0),  (0,  -  a,  0).  < 

6.  Plane  through  Three  Points.  Three  points,  not  lying  in  a 
straight  line,  determine  a  plane.  In  any  particular  case  the 
equation  of  the  plane  can  be  found  by  the  method  of  the  pre- 
ceding paragraph.  In  the  general  case,  when  the  points  are 
arbitrary  and  have  the  coordinates  (xly  yly  z^,  (a?2,  t/2,  22)> 
(xs>  2/3>  £3)5  this  method  could  still  be  applied.  It  is,  however, 
simpler  to  write  the  equation  in  determinant  form,  by  analogy 
to  the  equation  in  that  form  of  the  straight  line  through  two 
points  (Ch.  XVI,  §  11).  We  have,  namely, 

1 


(1) 


x      y 


=  0. 


Xi     y\     Zi 

#2        2/2       *2 
#3        2/3        *3 

To  show  that  this '  equation  actually  represents  the  plane 
through  the  three  points,  develop  the  determinant  by  the  minors 
of  the  first  row.  The  equation  then  takes  on  the  usual  form, 

Ax  +  By  +  Cz  +  D  =  0, 
of  a  linear  equation  ;  moreover,  the  values  obtained  for  A,  B,  C: 


c= 


2/2 
2/3 


are  not  all  zero,  since  otherwise  the  projections  of  the  three 
points  on  each  of  the  coordinate  planes  would  lie  on  a  line 
(Ch.  XVI,  §  11,  Th.  15)  *  and  hence  so  would  the  three  points 
themselves.  Consequently,  (1)  represents  some  plane. 

*  If  C,  for  example,  were  zero,  the  three  points  (Xi,  j/i,  0),  (X2,  y%,  0), 
(x3,  ^3,  0)  in  the  (x,  y)-plane  would  lie  on  a  line.  But  these  points  are 
the  projections  of  the  given  points  on  the  (x,  y)-plane. 


454  ANALYTIC   GEOMETRY 

This  is  the  plane  through  the  three  points,  since  the  sub- 
stitution of  the  coordinates  of  any  one  of  the  points  for  x,  y, 
z  in  (1)  makes  two  rows  of  the  determinant  identical  and  hence 
causes  the  determinant  to  vanish  and  the  equation  to  be 
satisfied. 

EXERCISES 

Find  the  equations  of  the  following  planes  by  applying 
formula  (1)  and  simplifying  the  result. 

1.  Through  (1,  2,  0),  (-  2,  3,  3),  (3,  -  1,  -  3). 

2.  Through  (2,  5,  -  3),  (-  2,  -  3,  5),  (5,  3,  -  3). 

3.  Through  (1,  1,  0),  (0,  1,  1),  (1,  0,  1). 

4.  Through  (1,  1,  -  1),  (6,  4,  -  5),  (-  4,  -  2,  3). 

5.  Through  (4,  5,  2),  (-  3,  -  2,  -  5),  the  origin. 

6.  Through  (xly  yl}  z^),  (xz,  yz,  z2),  the   origin.     Keep   the 
equation  in  determinant  form,  but  simplify  it. 

7.  Establish  the  intercept  form  of  the  equation  of  a  plane 
by  applying  formula  (1). 

8.  By  the   method   of  the   preceding   paragraph   find   the 
equation  of  the  plane  of 

(a)  Exercise  1  ;  (6)  Exercise  3  ;  (c)  Exercise  5. 

9.  Find  the  equations  of  the  faces  of  the  tetrahedron  whose 
vertices  are  at  the  points  (0,  0,  0),  (0,  3,  0),  (2,  1,  0),  (1,  1,  2). 

Ans.   z  =  0,  2x  =  z,  2x  —  4y+z  = 


7.   Perpendicular,  Parallel,  and  Identical  Planes.    Angle  be- 
tween Two  Planes.     The  normals  to  the  two  planes, 

(1)  AiX  +  Bfl  +  Cp  +  D^Q, 

(2)  A2x  +  B&  +  C2z  +  A  =  0, 

have  AI,  BI,  GI  and  A2,  B2,  Co,  respectively,  as  direction  com- 
ponents. 

The  planes  are  perpendicular  if  and  only  if  their  normals 
are  perpendicular;  and  parallel  (or  identical),  if  and  only  if 


THE    PLANE 


455 


the  normals  of  one  are  also  the  normals  of  the  other.     Conse- 
quently, we  have,  by  Ch.  XVIII,  §  4,  the  following  theorems. 

THEOREM  1.     The  planes  (1)  and  (2)  are  perpendicular  when 
and  only  ivhen 
(3). 

THEOREM  2. 


AiA*  +  B&  +  CiC8  =  0. 
The  planes  (1)  and  (2)  are  parallel  *  when  and 


only  when 

(4ij  j?±i :  x>i :  GI  =.42  :  _t>2  :  G2. 

The  condition  that  the  two  planes  be  identical  is  analogous 
to  the  condition  that  two  straight  lines  be  identical ;  cf.  Ch.  II, 
§  10.  We  can  state,  then,  the  theorem  : 

THEOREM  3.  The  planes  (1)  and  (2)  are  identical  when  and 
only  when 

(5)  A!  :  B!  :  Ci :  Dl  =  Az :  B2 :  C2 :  D2. 

The  proof  of  the  theorem  is  left  to  the  student. 

Angle  between  Two  Planes.  Between  the  two 
planes  (1)  and  (2)  there  are,  in  general,  two 
different  angles  having  values  between  0°  and 
180°  inclusive,  and  these  angles  are  supple- 
mentary. They  are  equal  to  the  angles  be- 
tween the  normals  to  the  two  planes.  Since 
AI,  jBt,  Ci  and  A2,  B2)  C2  are  the  direction  com- 
ponents of  the  normals,  the  cosines  of  the 
angles  are  given,  according  to  (3),  Ch.  XVIII, 
§4,  by 

A    A     i    T>  T>     i    n  n 

(6)  cos  0  =  ± 


FIG.  5 


+  A2  + 


+  BJ  +  C22 


EXERCISES 

In  each  of  the  following  exercises  determine  whether  the 
given  planes  are  parallel  or  perpendicular,  and  in  case  they 
are  neither,  find  the  angles  between  them. 

*  Or,  in  a  single  case,  identical.     Cf.  Th.  3. 


456  ANALYTIC   GEOMETRY 


0,  3z-?/-  92  +  2  =  0. 

2.  2x-y  +  3z-l  =  0,  2x  —  y  +  3z  +  3  =  0. 

3.  oj  +  ?/  +  z  =  0,  3z  +  6y-22  +  12  =  0 

4.  2<c-2.?/  +  42  +  5  =  0,  3z-3i/  +  62-l  =  0. 

5.  7z  +  5?/  +  62  +  30  =  0,  3x  —  y  -  Wz  +  4  =  0. 
6. 

7. 

8.  4z+'8y  +  2-8  =  0,  y  +  2-3  =  0. 

9.  2x  +  y  +  3z  —  2  =  Q,  x~2y  +  5  =  Q. 

10.  Show  that  two  planes  are  parallel  when  and  only  when 
their  equations  can  be  written  in  the  forms 

Ax  +  By  +  Cz  =  D,     Ax  +  By+Cz  =  D', 
D^D'. 

8.   Planes  Parallel  or  Perpendicular  to  a  Given  Plane. 

Example  1.  Find  the  equation  of  the  plane  which  passes 
through  the  point  (5,  2,  —  4)  and  is  parallel  to  the  plane 

(1)  2z  +  4?/-62-7  =  0. 

The  normals  to  the  plane  (1)  have  the  direction  components 
2,  4,  —  6  or,  more  simply,  1,  2,  —  3.  The  required  plane  has 
the  same  normals  and  passes  through  the  point  (5,  2,  —  4). 
By  (1),  §  2,  its  equation  is 

l(a;_5)+2(t/-2)-3(2*+4)=0, 
or  x  +  2y-3z-2l  =  0. 

Through  a  given  point  and  parallel  to  a  given  plane  there  is 
but  one  plane  ;  its  equation  can  always  be  found  by  the  above 
method.  But  'through  a  given  point  and  perpendicular  to  a 
given  plane  there  is  not  just  one  plane,  but  infinitely  many, 
namely,  all  the  planes  which  pass  through  that  normal  to  the 
given  plane  which  goes  through  the  given  point.  To  single 
out  one  of  these  planes  we  must  impose  a  further  condition. 
We  might  demand,  for  instance,  that  the  required  plane  pass 
through  a  second  given  point  or,  again,  we  might  specify  that 


THE   PLANE  457 

it  be  perpendicular  to  a  second  given  plane.  We  proceed  to 
consider  illustrative  examples  of  these  two  types. 

Example  2.  Find  the  equation  of  the  plane  passing  through 
the  two  points  (3,  —  2,  9),  (—  6,  0,  —  4)  and  perpendicular  to 
the  plane 

(2)  2x-y  +  4:z-8  =  0. 

First  Method.     Let  the  equation  of  the  plane  in  question  be 

(3)  Ax  +  By  +  Cz  +  D  =  0. 

Since  the  plane  contains  the  two  given  points,  we  must  have 

(4)  3A-2B  +  9C+D  =  0, 

(5)  -6A  -4O+Z>=0. 

Since  it  is  perpendicular  to  the  plane  (1),  it  is  necessary  that 

(6)  2^-5  +  4(7=0. 

In  (4),  (5),  (6)  we  have  three  simultaneous  linear  equations 
in  the  four  unknowns  A,  B,  C,  D.  But  from  three  linear 
equations  it  is  impossible  to  determine  uniquely  the  values  of 
four  unknowns.  It  may  be  possible,  however,  to  determine 
the  values  of  three  of  the  unknowns  in  terms  of  the  fourth, 
say  the  yalues  of  A,  B,  C  in  terms  of  D. 

Accordingly,  we  rewrite  the  equations  in  the  form 

3A-2B  +  9C  =  -  D, 
6  A  +4(7=     D, 

2A-    5  +  4(7=     0, 

and  solve  for  A,  B,  C,  either  directly  or  by  determinants.  We 
do  obtain  a  solution,  namely, 

A  =  \D,        B  =  -D,         C  =  -\D. 
Hence  (3)  becomes 

±Dx  -  Dy  -  %Dz  +  D  =  0. 

The  plane  represented  by  this  equation  is  always  the  same,  no 
matter  what  value,  other  than  zero,  is  given  to  D.  A  simple 
choice  is :  D=2.  We  obtain,  then,  as  the  equation  of  the 
required  plane, 

x-2    —  z  +  2  =  0. 


458  ANALYTIC   GEOMETRY 

Second  Method.  Let  N  be  any  normal  to  the  required  plane. 
Since  the  plane  contains  the  two  given  points,  N  is  perpen- 
dicular to  the  line  Lt  joining  these  points.  Since  the  plane  is 
perpendicular  to  the  given  plane  (2),  N  is  perpendicular  to 
any  line  L2  normal  to  (2).  Thus  N  is  a  common  perpendicular 
to  the  lines  LI  and  L*. 

The  direction  components  of  LI  are,  by  Ch.  XVIII,  §4, 
3_(_6),  -2-0,  9 -(-4),  i.e. 

9,     -2,     13; 
those  of  Lz  are  2,     —  1,       4. 

Consequently,  by  (6),  Ch.  XVIII,  §  5,  the  direction  components 
of  N  are 


-2    13 
-1      4 


13    9 
4    2 


9     -2 
2     -1 


i.e.  5,  -  10,  —  5,  or  1,  -  2,  -  1. 

Our  problem  is  now  reduced  to  that  of  finding  the  equation 
of  the  plane  which  passes  through  one  of  the  given  points, 
say  (3,  —  2,  9),  and  has  1,  —  2,  —  1  as  the  direction  com- 
ponents of  its  normals.  This  equation  is 

1(X  _  3)_  2(3,  +  2)-  l(z  -  9)=  0, 
or  x  —  2y  —  z  +  2=0. 

Example  3.  Find  the  equation  of  the  plane  passing  through 
the  point  (2,  5,  —  8)  and  perpendicular  to  each  of  the 
planes  : 


Either  of  the  methods  employed  in  the  previous  example  is 
applicable.  We  choose  the  latter.  A  normal  N  to  the  re- 
quired plane  is  perpendicular  to  the  normals  to  both  the  given 
planes.  These  have,  respectively,  the  direction  components 
2,  —  3,  4  and  4,  1,  —  2.  Consequently,  the  direction  compo- 
nents of  N  are  2,  20,  14  or  1,  10,  7. 

The  equation  of  the  plane  through  (2,  5,  —  8)  with  1,  10,  7 
as  the  direction  components  of  its  normals  is 


THE   PLANE  459 

(x  -  2)  +  10  (y  - 5 )  -f-  7(z  +  8)  =  0, 
or  x  +  10y  +  7z+-  4  =  0. 

This  is  the  required  equation. 

EXERCISES 

In  each  of  the  following  exercises  find  the  equation  of  the 
plane  which  is  parallel  to  the  given  plane  and  passes  through 
the  given  point.  In  Exs.  5,  6  find  the  equation  directly 
by  inspection  of  a  figure. 

Plane  Point 

1.  5x-2y  +  3z-4:  =  0,        (2,4,3). 

2.  3z  +  4?/-8z-2=0,  (0,0,0). 

3.  4z-2y-6z=9,  (2,  -  1,  0). 

4.  3z-4z  =  0,  (5,2,  -3). 

5.  3z  +  8  =  0,  (1,  -2,5). 

6.  2y-6  =  0,  (4,0,3). 

7.  Find  the  equation  of   the   plane   passing   through  the 
points   (3,  1,  2),    (3,  4,  4)   and   perpendicular   to   the   plane 
5x  +  y  +  4z=0.     Apply  both  methods,  checking  the  result  of 
one  by  that  of  the  other.  Ans.   2x  +  2y  —  3z  —  2  =  0. 

The  previous  problem,  if  the  given  points  and  the  given 
plane  are  as  specified.  Use  either  method  in  Exs.  8-10 ;  in 
Exs.  11,  12  solve  the  problem  directly  by  inspection  of  a 
figure. 

Points  Plane 

8.  (3,  4,  1),  (2,  6,  -  2),  2x-3y  +  4z-2  =  0. 

9.  (0,  0,  0),  (4,  3,  2),  x  +  y  +  z=0. 

10.  (3,  2,  -  4),  (5,  -  1,  3),        4  x  -  5y  =  8. 

11.  (1,  0,  0),  (1,  2,  5),  3y-7  =  0. 

12.  (0,  2,  0),  (2,  0,  0),  22  +  5  =  0. 

13.  There  are  infinitely  many  planes  which   pass  through 
the  two  points  (2,  —  3,  4),  (  —  2,  3,  —  6)  and  are  perpendicular 


460  ANALYTIC   GEOMETRY 

to  the  plane  whose  equation  is2a;  —  3t/  +  52  —  10  =  0.     Why  ? 
Justify  your  answer. 

14.  What  is  the  equation  of  the  plane  which  passes  through 
the  point  (1,  —  2,  1)  and  is  perpendicular  to  each  of  the 
planes  : 

—  2  =  0,        a; 


Apply  both  methods,  checking  the  result  of  one  by  that  of  the 
other.  Ans.   3x—2y—7z  =  Q. 

The  previous  problem  for  the  following  given  planes  and 
given  point.  In  Exs.  15-17  use  either  method  ;  in  Exs.  18,  19 
obtain  the  result  directly  from  a  figure. 

Planes  Point 


-  ,         ~    g  ™ 

•   (6*  +  2y-3*  +  4  =  0, 

16.    ja  +  y  +  *  =  °'  (1,  -1,1). 

17. 


18.  a?  =  2,  y  =  3,  (2,  -5,3). 

19.  2z  +  z  =  0,  3o;-2  =  6,         (2,  1,  -  3). 

20.  There  are  infinitely  many  planes  which  pass  through 
the  point  (2,  —  5,  0)  and  are   perpendicular   to   each  of   the 
planes  : 

±x-2y-6z  +  3  =  0,         -  6x  +  3y  +  92  +  10  =  0. 
Why  ?     Justify  your  answer. 

9.  Distance  of  a  Point  from  a  Plane.    To  find  the  distance 
A  of  the  point  P  :  (x0,  y0,  20)  from  the  plane 

Ax  +  By  +  Cz  +  D  =  0, 

draw  a  line  through  P  perpendicular  to  the  (a?,  y)-plane  and 
mark  the  point  Q  in  which  this  line  cuts  the  given  plane. 


THE   PLANE 


461 


Then,  as  the  figure  shows,  A  is  the  numerical  value  of   the 
product  QP  cos  6,  i.e. 

A  =  |  QPcosO\, 

where  0  is  the  acute  angle  between  the  line  QP  and  the  normal 
PP  to  the  plane. 

The  normal  P'P  has  the  direction 
components  A,  B,  C  and  the  line  QP 
has  the  direction  components  0,  0,  1. 
Consequently, 

C 

cos  0  =  ±     .  • 

V-42+  BP+  C2  /I? 

Fra.  6 
It  is  immaterial  to  us  which  of  the 

two  signs  is  the  proper  one,  for  we  are  interested  only  in 
numerical  values. 

It  remains  to  find  QP.  The  x-  and  y-coordinates  of  Q  are 
the  same  as  those  of  P,  namely,  x0,  y0 ;  denote  the  z-coordinate 
of  Q  by  zq.  Since  Q :  (XQ,  y0,  Zq}  lies  in  the  given  plane,  it  fol- 
lows that 

Ax0  +  By0  -f-  Czq  +  D  =  0, . 
and  hence  that 


Then 


or 


C 


_  Ax0  +  By0 


C 


Multiplying  the  values  obtained  for  QP  and  cos  9  together 
and  taktag  the  numerical  value  of  the  product,  we  obtain  the 
desired  formula : 


(1) 


+  52  +  (T2 


or 


=     Ax0 


462  ANALYTIC   GEOMETRY 

where,  in  the  second   formula,  the  sign  to  be   taken  is  that 
which  gives  a  positive  result. 

In  the  above  deduction  we  assumed  that  the  line  through  P 
perpendicular  to  the  (x,  y)-plane  meets  the  given  plane,  i.e.  that 
the  given  plane  is  not  perpendicular  to  the  (x,  #)-plane.  This 
means,  analytically,  that  we  have  assumed  that  (7^0. 

Since  we  know  that  at  least  one  of  the  three  coefficients 
A,  B,  C  is  not  zero  and  since  the  final  formula  (1)  bears  equally 
on  A,  B,  and  C,  it  is  immaterial  which  one  of  the  three  coeffi- 
cients we  assume  not  zero.  The  result  would  have  been  the 
same  if  we  had  assumed,  say,  A^Q,  instead  of  C  =£  0. 


EXERCISES 

1.  Establish  formula  (1)  on  the  assumption  that  A  =£  0. 

Find  the  distance  of  each  of  the  given  points  from  the  cor- 
responding given  plane.  In  Exs.  6,  7  check  the  result  by  in- 
spection of  a  figure. 

Point  Plane 

2.  (3,  -2,1),  2x-y  +  2z  +  3  =  0.  Ans.   4|. 

3.  (2,  5,  -3),  6z-32/  +  2z-4  =  0. 

4.  (0,2,1),  4x  +  3y  +  9  =  0. 

5.  Origin,  8x+y  —  4z  —  6  =  0. 

6.  (3,  8,  -  6),  y  -  5  =  0. 

7.  (-2,3,4),  20+7  =  0. 

8.  Find  the  lengths  of  the  altitudes  of  the  tetrahedron  of 
Ex.  9,  §  6. 

10.  Point  of  Intersection  of  Three  Planes.  Let  there  be 
given  three  planes  which  intersect  in  a  point,  i.e.  three  planes 
which  have  just  one  point  in  common,  as,  for  example,  the 
planes  of  the  ceiling  and  two  intersecting  walls  of  a  room,  or 
the  planes  of  three  faces  of  a  tetrahedron. 

The  point  of  intersection  of  the  planes  is  that  point  whose 
coordinates  satisfy  each  of  the  three  equations  of  the  planes. 


THE   PLANE  463 

In  other  words,  it  is  the  point  whose  coordinates  form  the 
simultaneous  solution  of  the  three  equations.  Consequently, 
to  find  its  coordinates  we  have  but  to  solve  the  three  equations 
simultaneously. 

Consider,  for  example,  the  three  planes  represented  by  the 
equations 

Sx  +  ly  -5z  =  -  11, 
2x+    y  +  6z  =  13, 
x-3y  +    2  =  6. 

The  simultaneous  solution  of  these  three  equations  is  most 
simply  effected  by  the  use  of  determinants  (Ch.  XVI,  §§2,  8). 
The  result  is  x  =  1,  y  =  —  1,  z  =  2.  Accordingly,  the  point  of 
intersection  of  the  three  planes  is  (1,  —  1,  2). 

Intersections  of  TJiree  Surfaces.  The  method  to  be  used 
in  finding  the  point  (or  points)  of  intersection  of  any  three 
surfaces,  given  by  their  equations,  is  now  obvious.  The  equa- 
tions are  to  be  regarded  as  simultaneous  equations  in  the  un- 
known quantities,  x,  y,  and  z,  and  solved  as  such. 

Three  Arbitrary  Planes.     Let  the  equations, 

A&  +  B,y  +  dz  +  A  =  0, 

(1)  A2x  +  B2y  +  C&  +  Dz  =  0, 

A3x  +  B3y  +  C\z  +  A  =  0, 

represent  three  distinct  planes.  If  the  determinant,  \A  B  C\, 
of  the  coefficients  of  x,  y,  and  z  is  not  zero,  the  three  equations 
have  a  unique  solution  (Ch.  XVI,  §  8,  Th.  10)  and  hence  the 
three  planes  intersect  in  a  single  point. 

Conversely,  if  the  three  planes  have  just  one  point  in  com- 
mon, \A  B  C |^=0.  For,  if  \A  B  C\  vanished,  the  normals  to 
the  three  planes  would  all  be  parallel  to  a  plane,  M,  by 
Ch.  XVIII,  §  6.  Consequently,  the  lines  of  intersection  of 
the  three  planes,  taken  in  pairs,  would  be  perpendicular  to  M. 
If  there  were  just  one  such  line  of  intersection,  the  three 
planes  would  have  all  the  points  of  this  line  in  common ;  if 
there  were  no  line  of  intersection  or  more  than  one,  the  three 


464  ANALYTIC   GEOMETRY 

planes  would  have  no  point  in  common.     In  either  case  the 
hypothesis  is  contradicted.     Hence  we  may  state  the  theorem : 

THEOKEM.  The  three  planes  (1)  intersect  in  a  single  point,  if 
and  only  if  \  A  B  C\  =£0.* 

If  |  A  B  C  |  =  0,  two  or  all  three  of  the  planes  may  be  parallel ; 
these  cases  are  easily  detected  by  inspection  of  the  equations. 
Or,  the  three  planes,  taken  in  pairs,  may  intersect  in  three 
distinct  parallel  lines.  Or,  finally,  they  may  have  a  line  in 
common.  We  shall  learn  later,  Ch.  XXI,  §  2,  how  to  dis- 
tinguish, from  the  equations  of  the  planes,  between  these  last 
two  cases. 

EXERCISES 

In  each  of  the  following  exercises  show  that  the  three  given 
planes  intersect  in  a  single  point,  and  find  the  coordinates  of 
the  point. 

1.  The  planes  of  Ch.  XVI,  §  2,  Ex.  10. 

2.  The  planes  of  Ch.  XVI,  §  2,  Ex.  11. 

3.  The  planes  of  Ch.  XVI,  §  2,  Ex.  12. 

4.  The  planes  of  Ch.  XVI,  §  2,  Ex.  13. 

5.  Find  the  coordinates  of  the  vertices  of  the  tetrahedron 
whose  faces  lie  in  the  planes 

z  =  0,     2y-3z  =  0,     x-y  +  3  =  0,     5x-2y  +  3z  =  0. 

Find  the  points  of  intersections  of  the  following  surfaces. 
Draw  a  figure  in  each  case. 

6.  x  —  4,     2  =  -  2,     xz  +  ?/2  =  25. 

7.  x  +  y  —  2,     x-y  =  Q,     a;2  +  z2— 1  =  0. 

8.  x2  -f  y2  +  z2  =  9,     5x  +  y  —  3z  =  5,     x  =  z. 

Ana.     (2,  1,  2),  (|,  I  I). 

*  Or,  the  three  equations  (1)  have  a  unique  solution,  if  and  only  if 
\A  B  C  |  =£  0.  This  is  the  converse  of  Th.  10,  Ch.  XVI,  §  8.  We  have 
thus  completed,  by  geometric  methods,  the  proof  of  an  important  fact  in 
the  theory  of  linear  equations. 


THE   PLANE  465 

In  each  of  the  exercises  that  follow  give  all  the  information 
you  can  concerning  the  relative  positions  of  the  three  given 
planes. 

Sx  —  4y-4z4-l  =  0,  9x  +  6y-3z  +  7  =  0, 

9.    —  2x  +    y+    2  +  5  =  0,      10.      x  —  2y  +    z  +  3  =  0, 
Qx-3y-3z-2  =  0.  6x  +  4y  -  2z  -  1  ==  0. 

11.  2x-3y  +  12  =  Q,    3x  +  5y-l  =  Q,    5o;+2y+ll  =  0. 

12.  4z  —  3z-5  =  0, 


EXERCISES  ON  CHAPTER  XIX 

1.  When  will  the  plane  Ax+By+Cz-\-D  =  0  pass  through 
a  coordinate  axis,  e.g.  the  axis  of  z  ?     When  will  it  be  parallel 
to  a  coordinate  axis,  e.g.  the  axis  of  x  ? 

2.  Find  the  equation  of  the  plane  through  the  axis  of  z 
and  the  point  (1,  2,  0). 

3.  Find  the  equation  of  the  plane  through  the  axis  of  y  and 
the  point  (2,  3,  1). 

4.  What  is  the  equation  of  the  plane  whose  intercepts  are 
one  half  those  of  the  plane  2  x  —  3  ?/  +  4  z  —  12  =  0? 

5.  A  perpendicular  from  the  origin  meets  a  plane  in  the 
point  (2,  —  3,  4).     What  is  the  equation  of  the  plane? 

6.  A  line  through  the  point  (2,  3,  7)  meets  a  plane  in  the 
point  (5,  —  1,  2).     Find  the  equation  of  the  plane. 

7.  Find  the  equation  of  the  plane  which  bisects  perpendic- 
ularly the  line  joining  the  points  (4,  3,  —  1),  (2,  5,  3). 

8.  Determine  the  point  on  the  axis  of  y  which  is  equidis- 
tant from  the  points  (3,  7,  4),  (—  1,  1,  2). 

9.  One  vertex  0  of  a  box  is  at  the  origin  and  the  edges 
issuing  from  Q  lie  along  the  positive  coordinate  axes.     Prove 
that  the  intercepts  of  the  plane  which  bisects  perpendicularly 
the   diagonal   through    0  are    inversely   proportional   to   the 
lengths  of  the  edges. 


466  ANALYTIC   GEOMETRY 

10.  For  what  value  of  m  will  the  two  planes, 

2x  +  my  —  2  =  4,        6x  —  5y  —  82  =  8, 
(a)  be  perpendicular  ?         (6)  be  parallel  ? 

11.  For  what  value  of  m  will  the  two  equations 

mx  —  y  +  z-\-3  =  Q,        4»—  my  +  raz  +  6  =  0 

represent  the  same  plane  ? 

12.  Find   the  angle   which   the   line   through    the    points 
(3,  2,  -  1),  (0,  4,  1)  makes  with  the  plane  2x-y-z  +  3  =  0. 

Suggestion.     Find  first  the  angle  between  the  line  and  a 
normal  to  the  plane. 

13.  What  angle  does  the  plane  3x  —  y  —  2  =  5  make  (a)  with 
the  (x,  y)-plane  ?     (&)  with  the  y-axis  ? 

14.  Find  the  distance  between  the  two  parallel  planes 


Suggestion.     Find  the  distance  of  a  chosen  point  of  the  first 
plane  from  the  second. 

15.    Show  that  the  distance  between  the  two  parallel  planes 
Ax  +  By  +  Cz  +  D  =  0,        Ax  +  By  +  Cz  +  D'  =  0 
IZX-DI 


IS 

V4^+W+C* 

16.  There  are  two  points  on  the  axis  of  z  which  are  distant 
four  units  from  the  plane  2  x  —  y  +  2z  +  3  =  Q.     Find  their 
coordinates.  Ans.  (0,  0,  4|),  (0,  0,  —  7|). 

17.  Show  that  the  equation  of  any  plane  parallel  to   the 
plane 

Ax  +  By  +  Cz  +  D=Q 

can  be  written  in  the  form 

Ax  +  By  +  Cz  =  k. 

18.  Using  the  method  of  Ch.  II,  §  11,  work  Exs.  1-4,  §  8,  of 
the  present  chapter. 


THE   PLANE 


467 


19.  There  are  two  planes  parallel  to  the  plane  2x—  6y+3  z 
=  4    and    distant    3    units    from    the    origin.      Find    their 
equations. 

20.  Find  the  equation  of   the  plane  parallel  to  the  plane 
given  in  Ex.  19  and  so  located  that  the  point  (3,  2,  8)  is  mid- 
way between  the  two  planes.         Ans.  2  x  —  6  y  -f-  3  z  —  32  =  0. 

21.  Three  faces   of   a   box   lie   in   the  planes  2x  —  y=6, 
x  -\-  2  y  —  8,  2  =  8  and  a  vertex  is  at  the  point  (9,  5,  2).     Find 
the  equations  of  the  planes  of  the  other  three  faces. 

22.  Find  the  equation  of  the  plane  which  passes  through 
the  point  (2,  —  1,  8)  and  is  parallel  to  each  of  two  lines  hav- 
ing 2,  —  3,  4  and  5,  —  7,  8  as  their  direction  components. 

Ans.    Ix  +  ly  +  z  —  12  =  0. 

23.  Show  that  the  equation  of  the  plane  passing  through 
the  points  (xl}  y^  z^),  (x»,  y2,  22)  an(i  perpendicular  to  the  plane 
Ax  +  By  +  Cz  +  D  =  0  can  be  written  in  the  form 


x      y      z      1 

2    y 2     2 
J.    JS    C    0 

24.    Show  that  the  four  planes, 

2  x  —  y  —  z  —  3  =  0,         x  - 


=  0. 


meet  in  a  point. 

25.  The  six  planes,  each  of  which  passes  through  the  mid- 
point of  an  edge  of  a  tetrahedron  and  is  perpendicular  to  the 
opposite  edge,  go  through  a  point. 

Prove  this  theorem  for  the  tetrahedron  of  Ex.  9,  §  6. 

26.  Prove  the  theorem  of  Ex.  25  for  the  general  tetrahedron, 
choosing  the  coordinate  axes  skillfully. 

27.  Show  that  the  plane 


468  ANALYTIC   GEOMETRY 

is  5  units  distant  from  the  origin  and  that  ^,  ^,  ^  are  the  di- 
rection cosines  of  a  normal  to  it,  directed  away  from  the 
origin. 

28.  State  and  prove  for  the  plane 

(1)  x  cos  a  +  y  cos  /3  +  z  cos  y  =  p,        p  >  0, 

the  results  corresponding  to  those  give"n  in  the  preceding  exer- 
cise. Show  that  the  equation  of  every  plane  can  be  written  in 
the  form  (1).  Prove,  also,  that  the  distance  of  the  point 
(xo>  yo>  2o)  from  the  plane  is 

|  x0  cos  «  +  y0  cos  ft  +  ZG  cos  y  —  P  }• 

29.  Prove  that,  if  a  plane  has  the  intercepts  a,  6,  c  and  is 
distant  p  units  from  the  origin, 


SYMMETRY 

30.  A  surface  is  symmetric  in  the  (x,  y)-plane  if  the  substitu- 
tion of  —  z  for  z  in  its  equation  leaves  the  equation  essentially 
unchanged.     Prove  this  theorem  and  state  the  corresponding 
theorems  for  symmetry  in  the  (y,  z)-  and  (z,  ar)-planes. 

31.  A  surface  is  symmetric  in  the  axis  of  z  if  the  substitution 
of  —  x  for  x,  and  of  —  y  for  y,  leaves  the  equation  essentially  un- 
changed.     Prove  this  theorem  and   state   the   corresponding 
theorems  for  symmetry  in  the  axes  of  x  and  y. 

32.  Prove  that  a  surface  is  symmetric  in  the  origin  if  the 
substitution  of  —  x  for  x,  of  —  y  for  y,  and  of  —  z  for  z,  leaves 
the  equation  essentially  unchanged. 

33.  Test  the  surfaces  of  the  following  exercises  of  §  1  for 
symmetry  in  each  coordinate  plane,  in  each  coordinate  axis,  and 
in  the  origin. 

(a)  Ex.  8  ;     (6)  Ex.  9  ;     (c)  Ex.  10  ;     (d)  Ex.  12. 

34.  Prove  the  following  theorems  : 


THE   PLANE  469 

(a)  If  a  surface  is  symmetric  in  each  of  two  coordinate 
planes,  it  is  symmetric  in  the  coordinate  axis  in  which  the  two 
planes  meet. 

(6)  If  a  surface  is  symmetric  in  each  coordinate  plane,  it  is 
symmetric  in  the  origin. 

(c)  If  a  surface  is  symmetric  in  a  coordinate  plane  and  in 
the  coordinate  axis  perpendicular  to  this  plane,  it  is  symmetric 
in  the  origin. 


CHAPTER  XX 
THE  STRAIGHT  LINE 

1.   Equations  of  a  Curve.    Example  1.     Given  the  two  equa- 
tions 

(1)  x  =  Q,  y  =  0. 

The  points  whose  coordinates  satisfy  both  equations  simultane- 
ously are  the  points  on  the  axis  of  z,  and  no  other  points.  We 
say  that  the  two  equations  represent  the  axis  of  z. 

Example  2.     Consider  the  two  equations 

(2)  Sx  —  ly  —  2  +  6  =  0,         5x  +  3y  +  22-8  =  0. 

A  point  whose  coordinates  satisfy  both  equations  at  once  must 
lie  in  each  of  the  two  planes  represented  by  the  equations,  i.e. 
it  must  be  a  point  on  the  line  of  intersection  of  these  planes. 
Conversely,  the  coordinates  of  any  point  on  this  line  satisfy 
both  equations.  Thus  the  two  equations,  considered  simultane- 
ously, represent  a  line,  the  line  of  intersection  of  the  two  planes 
which  the  two  equations,  taken  individually,  define. 

Example  3.     Take,  now,  the  pair  of  equations 

(3)  z2  +  /  +  z2  =  4,         x-y  =  0. 

By  reasoning  similar  to  that  of  Example  2,  it  follows  that 
these  equations,  taken  together,  represent  the  curve  of  inter- 
section of  the  two  surfaces  which  are  defined  by  the  two  equa- 
tions considered  individually.  The  first  equation  is  that  of  the 
sphere  whose  center  is  at  the  origin  and  whose  radius  is  two 
units  long.  The  second  equation  represents  the  plane  through 
the  axis  of  z  bisecting  the  angle  between  the  positive  x-  and  y- 

470 


THE   STRAIGHT   LINE 


471 


axes.  Consequently,  the  two  equations,  considered  simultane- 
ously, represent  the  circle  in  which  the  plane  intersects  the 
sphere. 

Example  4.     Consider,  lastly,  the  pair  of  equations 
(4)  x-2  +  y*  +  z2  =  4,         a;2  +(y  -  I)2  =  1. 

The  first  represents  the  sphere  of  Example  3.  The  second  is 
the  equation  of  the  circular  cylinder  erected  vertically  on  the 
circle  in  the  (x,  ?/)-plane  whose  center 
is  at  the  point  (0,  1,  0)  and  whose 
radius  is  unity.  The  curve  of  inter- 
section of  the  two  surfaces,  —  i.e.  the 
curve  represented  by  the  two  equa- 
tions taken  simultaneously,  —  is  shown 
in  Fig.  1.  It  does  not  lie  in  a  plane ; 
to  distinguish  it  from  curves  which 
do,  we  call  it  a  twisted  curve. 

Space  Curves.  The  straight  lines 
(1)  and  (2),  the  circle  (3),  and  the 
twisted  curve  (4)  are  all  called  space 
curves.  There  are,  then,  three  types 
of  space  curves:  straight  lines,  plane 
curves  other  than  straight  lines,  and 
twisted  curves. 

We  now  put  into  definitive  form 
what  we  have  learned  from  the  fore- 
going examples : 

Two  equations  in  x,  y,  z,  considered  simultaneously,  represent 
usually  *  a  space  curve.  The  curve  consists  of  all  those  points  and 
only  those  points  whose  coordinates  satisfy  simultaneously  both 
equations.  It  is  the  total  intersection  of  the  two  surfaces  which 
are  defined  by  the  two  equations  when  taken  individually. 

*  Two  equations  do  not  always  represent  a  curve.  For  example,  the 
pair  of  equations  x2  +  ?/2  =  0,  x2  +  22  =  0  represents  just  one  point,  the 
origin ;  and  the  equations  of  two  parallel  planes,  asx  —  y  +  z  =  2  and 
x  —  y  +  z  =  3,  represent  no  point  at  all  when  considered  simultaneously. 


FIG.  1 


472  ANALYTIC   GEOMETRY 

If  on  the  other  hand  it  is  a  curve,  and  not  a  pair  of  equations, 
which  is  given,  we  should  say : 

A  space  curve  can  be  represented  by  two  simultaneous  equations. 
These  can  be  ANY  two  equations  which  are  satisfied  simultaneously 
by  the  coordinates  of  every  point  of  the  curve  and  by  those  of  no 
other  point.  This  means,  geometrically,  that  the  curve  can  be 
considered  as  the  intersection  of  any  two  surfaces  on  which  it 
lies,  provided  the  two  surfaces  have  no  other  point  in  common. 

For  example,  a  straight  line  can  be  considered  as  the  inter- 
section of  any  two  planes  through  it.  Consequently,  it  can  be 
represented  by  the  equations  of  any  two  of  these  planes.  In 
other  words,  the  two  equations  of  the  line  are  not  unique.  Thus, 
as  equations  of  the  axis  of  z  we  might  take  the  equations  (1)  or 
we  might  take,  equally  well,  any  other  two  equations  represent- 
ing planes  through  the  axis  of  z,  as 


Equations  (1)  are,  however,  the  simplest  choice,  and  naturally 
we  shall  find  it  of  aid  in  analytical  work  to  choose  always  that 
pair  of  equations  representing  the  curve  under  consideration 
which  seems  most  simple. 

Problem  1.     What  are  the  equations  of  the  straight  line  L 
through  the  origin  with  the  direction  components  2,  3,  —  1  ? 

If  P:  (x,  y,  z)  is  any  point  on  the  line, 
other  than  the  origin  0,  then  x,  y,  z,  con- 
sidered as  the  projections  of  OP  on  the  axes, 
are  also  direction  components  of  L.  Hence, 
x,  y,  z  are  proportional  to  2,  3,  —  1 : 

44 

FIG.  2  (5)  x  —  2 py          y  —  3p}          z  =  —  p, 

where  the  factor  of  proportionality,  p,  is  not  constant,  but  vari- 
able, depending  for  its  value  on  the  position  of  P  on  the  line.* 
*  From  (5), 


V14 

That  is,  p  is  equal  numerically  to  the  distance  of  P  from  O,  divided  by 
V14.     Its  sign  depends  on  the  side  of  0  on  which  the  point  P  lies. 


THE   STRAIGHT   LINE  473 

Conversely,  if  P :  (x,  y,  z)  is  any  point  other  than  0,  whose 
coordinates  satisfy  equations  (5)  for  some  value  of  p,  these 
equations  say  that  the  direction  of  OP  is  that  of  L  and  hence 
that  P  lies  on  L. 

The  coordinates  (0,  0,  0)  of  the  excepted  point,  0,  obviously 
satisfy  (5) ,  when  p  is  given  the  value  0.  Consequently,  equa- 
tions (5)  represent  those  points  and  only  those  points  which 
lie  on  L,  i.e.  they  represent  L. 

Instead  of  equations  (5)  it  is  more  convenient  to  write : 

//*\  *^          V  * 

2   s    rr 

This  continued  equality  yields  the  three  equations : 
(7)  3«-2w  =  0,        y  +  3«  =  0,        x  +  2z  =  0. 

\       /  «7  7  «7        •  7 

One  of  these  equations  must  be  superfluous,  since  we  know 
that  two  equations  are  all  that  are  necessary  to  represent  a 
line.  As  a  matter  of  fact,  the  three  planes  defined  by  the 
three  individual  equations  all  pass  through  L  and  hence  one  of 
them  is  superfluous  in  determining  L.  We  prove  this  analyti- 
cally by  showing  that  a  simultaneous  solution  of  any  two  of 
the  three  equations  always  satisfies  the  third.  Thus,  if  x0,  y0, 
z0  are  any  values  of  x,  y,  z  which  satisfy  the  first  two  equations, 
i.e.  if 

3z0-2y0=0,        y0  +  3z0  =  0, 

elimination  of  y0  gives  the  relation 

x0  +  2  z0  =  0, 

which  says  that  these  values  also  satisfy  the  third  equation, 
q.  e.  d. 

Since  one  of  the  equations  (7)  is  superfluous,  we  might  take 
any  two  of  these  equations,  as 

to  represent  L.  It  is  more  convenient,  however,  to  consider 
the  continued  inequality  (6)  as  defining  L,  and  to  call  this 
continued  inequality  the  equations  of  L,  remembering  always, 
that  one  of  the  equations  which  follow  from  it  is  superfluous. 


474  ANALYTIC   GEOMETRY 

Problem  2.  Find  the  equations  of  the  curve  of  intersection 
C  of  two  circular  cylinders  of  the  same  radius  a,  whose  axes 
are  respectively  the  axes  of  x  and  y. 

The  equations  of  the  two  cylinders  are 

(8)  f  +  z2  =  a2,         x2  +  z2  =  a2, 

and  these  two  equations,  taken  together,  represent  the  curve  C. 
They  are  not,  however,  the  simplest  pair  of  equations  possi- 
ble.    If  x0)  2/0,  z0  are  any  set  of  values  of  x,  y,  z  satisfying  them 
simultaneously,  i.e.  if 

*/o2  +  *o2  =  «2,         Ob2  +  zo2  =  «2, 
elimination  of  z0  gives  the  relation 

^2-^  =  0, 

which  says  that  a^,  y0,  z0  also  satisfy  the  equation 

(9)  x*  -  y*  =  0. 

That  is,  the  curve  of  intersection  C  of  the  cylinders  (8)  lies  on 
the  surface  (9). 

Conversely,  the  surface  (9)  intersects  each  of  the  cylinders 
(8)  in  the  curve  C  and  in  no  other  points.  For,  if  x0,  y0,  20  are 
any  values  of  x,  y,  z  satisfying  equation  (9)  and  the  first,  say, 
of  equations  (8),  we  have  : 


Elimination  of  y0  gives  the  relation 


which  says  that  x0,  y0,  z0  satisfy  also  the  second  equation  of 
(8),  q.e.d. 

We  have  proved,  then,  that  the  total  intersection  of  any  two 
of  the  surfaces  (8)  and  (9)  is  the  curve  C.  Hence  any  two  of 
the  equations  (8)  and  (9)  define  C.  A  simpler  pair  than  the 
pair  (8)  is  the  combination  of  one  of  the  equations  (8)  with  the 
equation  (9),  for  example 

(10)  z2  +  z2  =  a2        a-2  -  f  =  0. 


THE   STRAIGHT    LINE 


475 


The  surface  (9)  consists  of  the  two  planes, 
x  —  y  =  0,         x  +  y  =  Q, 

passing  through  the  2-axis  and 
bisecting  the  angles  between 
the  (y,  z)-  and  (z,  #)-planes. 
Each  of  these  planes  intersects 
either  cylinder  in  an  ellipse ;  cf . 
Ch.  XII,  §  6.  Consequently, 
the  curve  C  consists  of  two  el- 
lipses. We  have,  then,  not  only 
obtained  the  simpler  equations 
(10)  to  represent  the  curve,  but 

we  have  also,  in  the  process,  succeeded  in   determining 
nature  of  the  curve. 

EXERCISES 

In  each  of  the  following  exercises  determine  what  the  given 
equations,  considered  simultaneously,  represent.  Draw  a 
figure. 


FIG.  3 


the 


1.  y  =  0,2  =  0. 

3.  x  +  4  =  0,  z  —  3  =  0. 

5.  y  —  z  =  0,  x  =  3. 

7.  y  +  2  =  2,  2  x  =5. 


9.    -  =  ^  =  -- 
326 


2.  x  =  4,  y  =  0. 

4.  2  y+  3  =  0,3  z-5=0. 

6.  3x  +  2y=Q,  2-4=0. 

8.  x  —  y  =  0,  x  —  z  =  0. 

10.  x  =  —  y  =  2. 


11.  x2  +  y2  +  z2  —  9  =  0,  y  —  2. 

12.  x2  +  yz  +  z2  —  16  =  0,  x  +  z  =  4. 

13.  x1  +  yz  =  4,  2  y  +  z  =  3. 

14.  x2  +  y*  +  z2  —  25  =  0,  x2  +  y2  =  16. 

15.  a2  +  ?/2  +  z2  -  16  =  0,  4  z2  +  (y  —  2)2  =  4. 

16.  xz  +  y2  =  a2,  z2  =  ay. 

17.  Which  of  the  curves  represented  by  the  above  pairs  of 
equations  pass  through  the  origin  ? 


476  ANALYTIC    GEOMETRY 

Find  the  equations  of  the  following  curves : 

18.  The  axis  of  y. 

19.  The  line   in   the  (x,  y)-plane  3  units  in. front  of   the 
(y,  z)-plane. 

20.  The  line  2  units  to  the  left  of  the  (z,  x)-plane  and  3 
units  above  the  (x,  y)-plane. 

21.  The  line  f  units  behind  the  (y,  z)-plane  and  £  units  to 
the  right  of  the  (z,  ce)-plane. 

22.  The  line  which  lies  in  the  plane  passing  through  the 
ovaxis  and  bisecting  the  angle  between  the  positive  y-  and 
z-axes  and  is  4  units  above  the  (x,  y)-plane. 

23.  The   line   through  the  origin  with   the  direction   com- 
ponents 1,  —  1,  —  1. 

24.  The  line  through   the   origin  with  the  direction   com- 
ponents 2,  0,  3. 

25.  The  circle  of  radius  3  whose  center  is  on  the  axis  of  z 
and  whose  plane  is  4  units  above  the  (x,  2/)-plane. 

26.  The   circle  of   radius  2  whose  center  is  at  the  origin 
and  whose  plane  passes  through  the  y-axis  and  bisects  the 
angle  between  the  positive  axis  of  x  and  the  negative  axis  of  z. 

Find  a  simpler  pair  of  equations  to  represent  the  curve 
given  in  each  of  the  following  exercises  and  then  identify  the 
curve.  Draw  a  figure. 

27.  2  y  -  3  z  =  0,  5  y  +  16  z  =  0. 

28.  x  +  y  +  z  =  2,  x  +  y  —  5  =  0. 

29.  The  curve  of  Ex.  14. 

30.  or2  +  ?/2  +  z2  =  a2,  2  yz  +  z2  =  a2. 

2.   Line  of  Intersection  of  Two  Planes.     Let  the  plane 
and  the  plane 


THE   STRAIGHT   LINE 


477 


be  two  planes  which  meet.  According  to  the  theory  of  the 
preceding  paragraph,  their  line  of  intersection  is  represented 
by  the  two  simultaneous  equations 


A2x  +  B2y+C2z  +  D2  =  0, 

i.e.  by  these  equations,  where  the  only  sets  of  values  of  x,  y, 
z  considered  are  those  which  satisfy  both  equations. 

Since  the  line  lies  in  each  plane,  it  is  perpendicular  to  the 
normals  to  each  plane.  That  is,  it  is  a  common  perpendicular 
to  the  normals  to  the  two  planes.  From  this  fact  its  direc- 
tion components  can  easily  be  determined,  by  the  method  of 
Ch.  XVIII,  §  5. 

Consider,  for  example,  the  line  (2)  of  §  1.  The  normals  to 
the  two  planes  determining  this  line  have  3,  —  4,  —  1  and  5, 
3,  2  as  direction  components.  Consequently, 


-4 
3 


-1 

2 


-1 

2 


3     -4 
5         3 


or  —  5,  —  11,  29  are  the  direction  components  of  the  line. 

The  planes  determining  the  line  (1)  have  normals  with  AI, 
BI,  Ci  and  A«,  B2,  Q  as  direction  components.  Hence,  the 
direction  components  of  the  line  (1)  are 


(2) 


-B 


d 
C2 


A, 


A, 


Parallel  Planes.  If  two  planes  are  parallel  and  so  have  no 
line  of  intersection,  their  equations  are  incompatible,  i.e.  they 
have  no  simultaneous  solution.  For  example,  the  equations 

2z  +  5  =  0, 


which  represent  two  parallel  planes,  are  obviously  incompati- 
ble, since  the  first  says  that  the  quantity  2x  +  y  —  z  has  the 
value  3,  whereas  the  second  says  that  it  has  the  value  —  f  . 


478 


ANALYTIC   GEOMETRY 


EXERCISES 

Find  the  direction  components  of  each  of  the  following 
lines.  Determine  also  a  point  on  the  line  and  hence  construct 
the  line. 


2.  3x-±y-6z  +  7  =0, 

3.  x  +  3y  —  3z  +  5=Q, 

4.  2x  —  y  —  z  =  0, 

5.  3x-5y-z-l  =  0, 

6.  x  +  2     =  0 


2x-  y-  2z  +  1  =  0. 
3x  +  ±y  +  62  -  5  =  0. 

x  4-  y  =  5. 
x  —  4  =  0. 
y-3z  =  0. 


FIG.  4 


7.  What  are  the  equations  of  the  edges  of  the  tetrahedron 
of  Ch.  XIX,  §  6,  Ex.  9  ?     Draw  a  figure  and  label  the  edges 
and  the  pairs  of  equations  to  correspond. 

8.  The  same  for  the  tetrahedron  of  Ch.  XIX,  §  10,  Ex.  5. 

9.    Show  that   the   lines  of  Exs.  1 
and  4  are  perpendicular. 

10.  Show  that   the  lines   of  Exs.  3 
and  6  are  parallel. 

11.  Each  side  of  a  hip  roof  (Fig.  4) 
makes  an  angle  with  the  horizontal  whose  tangent  is  ^.     What 
angle  do  the  edges  of  the  roof  make  with  the  horizontal  ? 

3.  Line  through  a  Point  with  Given  Direction  Components. 

Let  it  be  required  to  find  the  equations  of 
the  line  L  which  goes  through  the  point 
P0  :  (x0,  ?/0,  z0)  and  has  the  direction  com- 
ponents I,  m,  n. 

Let  P  :  (x,  y,  z)  be  any  point  on  L,  other 
than  P0.  Then  the  direction  components 
of  P0P,  namely, 


(1) 


THE    STRAIGHT    LINE  479 

are  direction  components  of  L.  They  are,  then,  proportional 
to  I,  m,  n: 

/O\  7 

(—  )  *C  "~~  «*/Q  ^— •  P^J  y  """""  jO  """"   P        9  ~~~       0  ~"~~  P      9 

where  the  factor  of  proportionality,  p,  varies  in  value  as  P 
changes  position.* 

Conversely,  if  P :  (x,  y,  z)  is  any  point  other  than  P0,  for 
which  equations  (2)  hold  for  some  value  of  p,  i.e.  for  which 
the  quantities  (1)  are  proportional  to  Z,  m,  n,  it  follows  that 
the  direction  of  P0P  is  that  of  L  and  hence  that  P  lies  on  L. 

The  coordinates  (x0,  y0,  z0)  of  the  excepted  point  P0  obviously 
satisfy  (2),  when  p  =  0.  Consequently,  equations  (2)  are  satis- 
fied by  the  coordinates  of  those  points  and  only  those  points 
which  lie  on  L  and  so  they  represent  L. 

Instead  of  equations  (2),  we  can  write 

/o\  *B     «BQ y    yp 3  —  ZQ  ^ 

I  m  n 

We  calk  this  continued  equality  the  equations  of  the  line, 
remembering  from  §  1  that,  of  the  three  equations  which  in 
general  result  from  it,  one  is  superfluous. 

If  one  of  the  denominators  in  (3)  is  zero,  so  is  the  cor- 
responding numerator ;  thus,  if  w  =  0,  then  z  —  z0  =  0.  For, 
equations  (3)  are  but  an  abbreviated  form  of  the  equations  of 
proportionality  f  (2),  and  if  n  =  0  the  last  equation  in  (2) 
reduces  immediately  to  z  —  z0  =  0. 

Suppose,  for  example,  that  the  line  is  to  go  through  the 
point  (3,  4,  —  6)  and  have  0,  5,  3  as  its  direction  components. 
Here  I  =  0,  and  hence  x  —  x0  =  0,  i.e.  x  —  3  =  0.  Thus, 

*  From  (2),  P0P2  =  p2  (P  +  m?  +  n»), 

PoP 


or  p  =  ±      

Vl2  +  m2  +  n2 

That  is,  the  numerical  value  of  p  is  proportional  always  to  the  distance 
of  P  from  P0.  The  sign  of  p  depends  on  the  side  of  P0  on  which  the 
point  P  is  situated. 

t  Cf.  Ch.  XVI,  §  9,  eq.  (5). 


480  ANALYTIC   GEOMETRY 

x  —  3  =  0  is  one  of  the  equations  of  the  line.     The  other, 
obtained  from  the  equality  of  the  last  two  members  of  (3),  is 


or 


5  3 

Hence  the  desired  equations  are 

x  -3  =  0,        3  z-5  z-42  =  0. 

Again,  let  the  point  be  (1,  2,  —  3)  arid  let  the  direction  com- 
ponents be  0,  0,  .1.  In  this  case,  I  =  0,  and  m  =  0  ;  therefore 
x  —  x0  =  0  and  y  —  y0  =  0,  i.e. 

x  —  1  =  0,         y-2  =  0, 

and  these  are  the  equations  of  the  line.  This  result  might 
have  been  obtained  directly,  by  inspection,  since  it  is  clear 
that  the  line  is  parallel  to  the  axis  of  z. 

Reduction  of  a  Continued  Equality  to  the  Form  (3).  Ex- 
ample 1.  What  are  the  direction  components  of  the  line 

x  —  5  _     y  -f  4_z  +  39 
6  32 

This  continued  equality  will  be  of  the  form  (3),  if  the  minus 
sign  before  the  second  fraction  is  associated  with  the  denomi- 
nator. Accordingly,  the  direction  components  are  6,  —  3,  2. 

Example  2.     Consider  a  more  complicated  case  : 

2x-l=2-5y  =  3z 
3  4  2  ' 

To  put  this  continued  equality  in  the  form  (3),  divide  the 
numerator  and  denominator  of  each  fraction  by  the  coefficient 
of  the  variable  in  the  numerator  : 


I          -*      * 

Here,  each  variable  x,  y,  z,  as  it  appears  in  the  numerator,  has 
the  coefficient  unity,  and  there  is  complete  conformity  to  (3). 


THE   STRAIGHT   LINE  481 

The  direction  components  of  the  line  are,  then,  |  ,  —  -|,  f  or 
45,  —  24,  20.  Furthermore,  it  is  clear  that  the  line  goes 
through  the  point  (-^,  f  ,  0). 

EXERCISES 

In  each  of  the  following  exercises  find  the  equations  of  the 
line  through  the  given  point  with  the  given  direction  com- 
ponents. 

Point  Components 

1.  (2,  -3,1),  5,2,  -4. 

2.  (0,  0,  0),  3,  -  1,  2. 

3.  (4,  -1,  -2),  -6,5,8. 

4.  (2,0,  -3),  1,1,1. 

5.  (3,2,  -8),  1,3,0. 

6.  (2,0,1),  4,0,1. 

7.  (-3,4,6),  0,1,0. 

In  each  of  the  exercises  which  follow,  find  the  direction 
components  of  the  line  represented  by  the  given  equations 
and  the  coordinates  of  a  point  on  the  line.  Construct  the  line. 


3 


=    y   =  2z-l 


9-64 

10.  1  —  x  =  y  —  2  =  z  —  6. 

11.  3x+  4  =  2  —  5y  =  4:Z  —  7. 

12.  2x  =  l  —  y  =  3z. 

13.  Show  that  the  lines  of  Exs.  8  and  10  are  perpendicular. 

14.  Show  that  the  lines  of  Exs.  9  and  12  are  parallel. 

4.   Line  through  Two  Points.     The   line  through  the  two 
points  (xi,  yi,  z^),  (#2,  y^  z2)  has  the  direction  components 

«a  -  *i>        2/2  -  2/i,        *2  ~  zi- 


482  ANALYTIC  GEOMETRY 

It  can  be  considered,  then,  as  the  line  which  has  these  direction 
components  and  goes  through  the  point  (xl}  2/1,  Zi).  Conse- 
quently, by  (2),  §  3,  it  is  represented  by  the  equations 

(1)    x  —  x1  =  P(x2-x1'))  2/  -  2/1  =  /»0/2  -  2/i)>    z-zi 


Instead  of  (1)  we  can  write,  as  the  equations  of  the  line,  the 
continued  equality 

(2)  x  —  Xj  _  y  —  yl  _  z  —  z^ 

®2  -  KI     ?/2-2/i     z2  -  »i 

Since  (2)  is  an  abbreviated  form  of  (1),  it  follows  that,  if  a 
denominator  in  (2)  is  zero,  the  corresponding  numerator  is  also 
zero.  '  Thus,  if  the  two  points  are  (3,  5,  —  4),  (8,  5,  —  4),  so 
that  2/2  —  2/1  =  0  an(i  z2  —  zl  =  0,  we  have  y  —  yl  =  0  and 
z  —  Zi  =  0,  that  is, 

2,  _  5  =  0,         2  +  4  =  0. 

These  are,  then,  the  equations  of  the  line.  The  result  might 
have  been  obtained  directly  by  noting  in  the  beginning  that 
the  2/-coordinates,  and  also  the  z-coordinates,  of  the  two  points 
are  equal  and  by  concluding,  then,  that  the  line  is  parallel  to 
the  axis  of  x. 

EXERCISES 

Find  the  equations  of  each  of  the  following  lines. 

1.  Through  (2,  5,  8),  (-  1,  6,  3). 

2.  Through  (-  1,  0,  2),  (3,  4,  6). 

3.  Through  the  origin  and  (5,  —  2,  3). 

4.  Through  (2,  0,  3),  (0,  3,  2).  ' 

5.  Through  (2,  -  5,  8),  (2,  3,  7). 

6.  Through  (3,  -  2,  -  5),  (3,  -  2,  6). 

7.  The  edges  of  the  tetrahedron  of  Ch.  XIX,  §  6,  Ex.  9. 

8.  The  edge  of  the  quadrangular  pyramid  of  Ch.  XIX,  §  5, 
Ex.  10. 


THE   STRAIGHT   LINE  483 

5.  Line  or  Plane  in  Given  Relationship  to  Given  Lines  or 
Planes.  Problem  1.  To  find  the  equations  of  the  line  through 
a  given  point  perpendicular  to  a  given  plane. 

The  direction  components  of  the  line  are  those  of  a  normal  to 
the  given  plane  and  hence  can  easily  be  found.  The  problem 
then  becomes  that  of  finding  the  equations  of  a  line  through  a 
given  point  with  given  direction  components. 

For  example,  if  the  given  point  is  the  origin  and  the  given 
plane  is 

3x  —  4y  +  5z  +  6  =  0, 

the  required  line  goes  through  (0,  0,  0)  and  has  the  direction 
components  3,  —  4,  5.     Hence  its  equations  are 

x—0_y— 0=z— 0 

3          -4  =      5 
or  20x  =  -15y  =  12z. 

Problem  2.  To  find  the  equation  of  the  plane  through  a  given 
point  perpendicular  to  a  given  line. 

The  direction  of  a  normal  to  the  plane  is  that  of  the  given  line 
and  therefore  the  direction  components  of  the  normal  are  easily 
written  down.  We  then  have  the  problem  of  finding  the  equa- 
tion of  a  plane  through  a  given  point  with  given  direction  com- 
ponents of  its  normals. 

Thus,  if  the  point  is  (3,  —  2,  1)  and  the  line  is  given.,  as  the 
intersection  of  two  planes,  by  the  equations, 

3x-5y-2z  +  6  =  Q,        4x  +  y  +  3z  —  1  =  0, 
the  direction  components  of  the  line  are,  by  (2),  §  2, 


-5     -2 
1         3 


-2    3 
3    4 


3  -5 

4  1 


i.e.  —  13,  —  17,  23,  or  13,  17,  —  23.  The  required  plane  has 
these  direction  components  for  its  normals  and  passes  through 
the  point  (3,  —  2,  1).  Consequently,  its  equation  is 

13(a;  -  3)  +  17 (y  +  2)  -  23(z  -  1)=  0, 
or 


484  ANALYTIC   GEOMETRY 

Problem  3.  Let  two  non-parallel  lines,  LI  and  Z/2,  and  a  point 
P  be  given.  Through  P  parallel  to  each  of  the  lines  there  is  a 
unique  plane.  To  determine  this  plane. 

A  normal  to  the  plane  is  a  common  perpendicular  to  the  lines 
LI  and  Z/2  an(i  so  its  direction  components  can  be  found  by  the 
method  of  Ch.  XVIII,  §  5.  The  problem  is  then  that  of  finding 
the  equation  of  a  plane  through  a  given  point  with  given  direc- 
tion components  of  its  normals. 

If  LI  and  Z/2  have  the  equations 

x—  1_2/  —  2 z          #  +  3__y_z  —  1 

5  3          ~2'         ~~4~~  ~2~     3     ' 

their  direction  components  are,  respectively,  5,  3,  —  2  and  4, 2, 
3.  The  direction  components  of  a  common  perpendicular  to 
them  are,  by  Ch.  XVIII,  §  5,  (6),  13,  -  23,  -  2.  Thus  the 
plane  parallel  to  LI  and  Z/2  and  passing  through  a  given  point, 
say  (3,  2,  —  4),  has  the  equation 

13  (x  -  3)  -  23  (y  -  2)  -  2  (z  +  4)  =  0, 
or  13ce-23t/-2z-l  =  0. 

Problem  4.  Given  two  intersecting  planes,  Ml  and  M2,  and 
a  point  P.  Through  P  parallel  to  each  of  the  planes  there  is  a 
unique  line.  To  find  this  line. 

Since  the  line  is  parallel  to  each  of  the  planes,  it  is  parallel 
to  their  line  of  intersection.  It  is,  therefore,  itself  the  line  of 
intersection  of  the  two  planes  which  pass  through  P  and  are 
parallel  respectively  to  Ml  and  Mz. 

For  example,  if  P  is  (2,  0,  —  1)  and  Mv  and  Mz  are 

2x-3y  +  z-  6  =  0,         4z  -  2y  +  3z  +  9  =  0, 

the  planes  through  P  parallel  respectively  to  J/i  and  M2  have 
the  equations 

2(aj-  2)-  3y  +  (z  +  1)=  0,     4(«  -  2)-  2y  +  3(z  +  1)=  0, 
or         2a;  — 3y  +  2  —  3  =  0,         4z  -  2y  +  3z  -  5  =  0. 

These  equations,  considered  simultaneously,  represent  the  re- 
quired line. 


THE    STRAIGHT   LINE  485 

The  problem  might  have  been  solved  by  determining  the  di- 
rection components  of  the  line  of  intersection  of  Ml  and  M2,  as 
given  by  (2),  §  2,  and  by  finding  the  equations  of  the  line  which 
has  these  direction  components  and  passes  through  P. 

Problem  5.  To  find  the  equation  of  a  line  which  passes 
through  a  given  point  and  is  parallel  to  a  given  line. 

If  the  line  is  given  as  the  intersection  of  two  planes,  this  is 
the  previous  problem.  If  its  equations  are  given  in  the  form  of 
a  continued  equality,  the  solution  is  simple.  We  leave  it  to  the 
student. 

Problem  6.  To  find  the  equations  of  a  line  which  passes 
through  a  given  point  and  is  perpendicular  to  each  of  two  given 
non-parallel  lines. 

The  solution  of  this  problem  we  also  leave  to  the  student. 

EXERCISES 

In  each  one  of  the  following  exercises  in  which  it  is  possible, 
solve  the  given  problem  directly,  by  inspection  of  a  figure. 

Find  the  equations  of  the  line  passing  through  the  given 
point  and  perpendicular  to  the  given  plane. 

Point  Plane 

1.    (2,  -  8,  3), 
2-    (0,0,0), 

3.  (3,  -4,0), 

4.  (-1,2,5),  22  +  3  =  0. 

Find  the  equation  of  the  plane  passing  through  the  given 
point  and  perpendicular  to  the  given  line. 

Point  Line 

2x-3y  +  6z-4=  =  Q, 


6 
6< 


486  ANALYTIC   GEOMETRY 

Point  Line 

7.   (3,  -2,  -8),  «=l___*_r 


8.  (6,  0,.-1), 

9.  (4,  -i,3),- 

Find  the  equation  of  the  plane  passing  through  the  given 
point  and  parallel  to  each  of  the  given  lines. 

Point  Lines 

10.  (3,  5,  1)  The  lines  given  in  Exs.  7,  8. 

11.  (0,  2,  -  3)  The  lines  given  in  Exs.  6,  7. 

12.  (0,  0,  0)  The  lines  given  in  Exs.  5,  6. 

13.  (2,  -|,3)  ^  =  4,3  =  3;  2*  =  5,32/  =  7. 

Find  the  equations  of  the  line  passing  through  the  given 
point  and  parallel  to  each  of  the  given  planes. 

Point  Planes 

14.  (0,  0,  0)  The  planes  given  in  Exs.  1,  2. 

15.  (2,  —  1,  —  3)  The  planes  given  in  Exs.  1,  3. 

16.  (0,  2,  —  f)  The  planes  given  in  Exs.  2,  3. 

17.  (|,  0,  —  |)  The  planes  given  in  Exs.  3,  4. 

Find  the  equations  of  the  line  passing  through  the  given 
point  and  parallel  to  the  given  line. 

18.  The  point  and  line  given  in  Ex.  5. 

19.  The  point  and  line  given  in  Ex  7. 

20.  The  point  and  line  given  in  Ex.  8. 

21.  The  point  and  line  given  in  Ex.  9. 

Find  the  equations  of  the  line  passing  through  the  given 
point  and  perpendicular  to  the  given  lines. 

Point  Lines 

22.  (0,  0,  0)  The  lines  given  in  Exs.  7,  8. 

23.  (3,  -  2,  5)  The  lines  given  in  Exs.  6,  7. 


THE   STRAIGHT   LINE  487 

Point  Lines 

24.  (2,  4,  0)  The  lines  given  in  Exs.  5,  6. 

25.  (0,  0,  3)]  The  lines  given  in  Ex.  13. 

6.  Angle  between  a  Line  and  a  Plane.  Given  a  plane  M 
and  a  line  L  which  is  not  perpendicular  to  the  plane.  Pro- 
ject the  line  on  the  plane.  The  acute 
angle,  <£,  which  the  line  makes  with  this 
projection  is  the  angle  between  the  line 
and  the  plane.  It  may  be  determined  /M 
by  finding  the  acute  angle,  0,  which  the  FlG  6 

line  makes  with  a  normal  N  to  the  plane. 
For,  it  is  clear  that  6  and  <f>  are  complementary  angles. 

Example.     Find  the  angle  between  the  line 

2z  +  2  =  2/  +  l=-4 
and  the  plane 


The  direction  components  of  the  line  are  2,  4,  —  1  ;  those  of 
a  normal  to  the  plane  are  2,  3,  —  2.     Hence 


V4  +  16  +  1V4  + 

where  we  are  to  take  that  sign  which  makes  the  right-hand 
side  positive. 

Thus,         cos  0  =  -^-  =  0.9527.    and     0  =  17°  42'. 

V357 

Finally,  <£  =  90°  -  0  =  72°  18'. 

EXERCISES 

Find  the  angle  between  the  given  line  and  the  given  plane. 

Line  Plane 

1.  3aj  +  3  =  2y  +  2=-6«  —  12,     3x  +  y  +  2z  +  1  =0. 

2.  f3*-42,  +  2*  =  0,  3x-22-12  =  0. 

14#  —  3y  +  z  =  5, 


488  ANALYTIC   GEOMETRY 

3.  Through  (0,  0,  0),  (1,  2,  -  2),      2x  -  6y  +  3z  -  4  =  0. 

4.  Through  (2,  5,  3),  (4,  -  1,  6),     y  +  2  z  +  4  =  0. 

5.  Find  the  angles  which  the  plane  given  in  Ex.  1  makes 
with  the  coordinate  axes. 

6.  Find  the  angles  which  the  line  given  in  Ex.  2  makes 
with  the  coordinate  planes. 

7.  Point  of  Intersection  of  a  Line  and  a  Plane.    Given  a  plane, 

(1)  Ax  +  By  +  Cz  +  D  =  0, 
and  a  line, 

(2)  AIX  +  B&  +  C&  +  A  =  0,    A&  +  B&  +  C2z  +  D2  =  Q, 

which  intersects  the  plane.  The  coordinates  of  the  point  of 
intersection  satisfy  the  equation  of  the  plane,  since  the  point 
lies  in  the  plane.  They  also  satisfy  each  of  the  equations  of 
the  line,  for  the  point  lies  on  the  line.  They  are,  then,  the 
simultaneous  solution  of  the  three  equations 

Ax+By  +  Gz  +  D  =  0, 

(3)  Ap  +  Btf  +  Ci*  +  Dj  =  0, 

=  0. 


Example.  Find  the  coordinates  of  the  point  of  intersection 
of  the  line  and  the  plane  of  §  6.  Take,  as  the  equations  of 
the  line,  those  obtained  by  equating  the  first  and  second  mem- 
bers, and  then  the  first  and  third  members,  of  the  continued 
equality  which  represents  the  line.  Then  the  three  equations, 
which  are  to  be  solved  simultaneously,  are 


The  solution  is  found  to  be  x  =  1,  y  =  3,  z  =  6.     Thus,  the 
line  meets  the  plane  in  the  point  (1,  3,  6). 

Intersection  of  a  Curve  and  a  Surface.  The  above  method 
applies  also  to  the  problem  of  finding  the  point  (or  points) 
of  intersection  of  a  curve  and  a  surface  which  are  given  by 
their  equations.  The  two  equations  of  the  curve  and  the 


THE    STRAIGHT    LINE  489 

equation  of  the  surface  are  to  be  considered  as  simultaneous 
equations  in  the  unknowns,  x,  y,  z,  and  solved  as  such. 

Plane  and  Line  Arbitrary.  The  line  (2)  has  one  and  just 
one  point  in  common  with  the  plane  (1),  if  and  only  if  the 
three  planes  (3)  intersect  in  a  single  point,  i.e.  by  the  theorem 
of  Ch.  XXIX,  §  10,  if  and  only  if  the  determinant  of  the  coeffi- 
cients of  x,  y,  z  in  equations  (3)  does  not  vanish. 

If  this  determinant  vanishes,  it  follows,  either  directly  or 
from  the  discussion  in  Ch.  XXIX,  §  10,  that  the  line  either  is 
parallel  to  the  plane  or  lies  in  it. 

EXERCISES 

•  . 

Show  that  the  given  line  has  just  one  point  in  common  with 

the  given  plane  and  find  the  coordinates  of  the  point. 

1.  The  line  and  plane  of  Ex.  1,  §  6. 

2.  The  line  and  plane  of  Ex.  2,  §  6. 

3.  The  line  and  plane  of  Ex.  3,  §  6. 

Find  the  points  of  intersection  of  the  given  curve  with  the 
given  surface.  Draw  a  figure  for  each  exercise. 

Curve  Surface 

4.  x  =  y  =  z,  xt  +  y*  +  z*  =  l. 

5.  z2  +  ?/2  +  z2  =  29,  z  =  2,  4:X-3y  =  0. 

6.  12-6o;  =  2?/  +  2  =  32-9, 

7.  a;5  +  02  +  «2  =  12,  x  =  y, 


Find  out  all  you  can  about  the  relative  positions  of  the  given 
line  and  the  giyen  plane. 

Line  Plane 

8.  " 

9.  x  =  y  =  z.  5x  +  3y  —  8z  —  3  =  0. 

I  2x  +  3y-  8  =  0, 
10.    \  9  8x~y—6  =  0, 


490  ANALYTIC   GEOMETRY 

8.  Parametric  Representation  of  a  Curve.  The  Straight  Line. 
Given  a  directed  straight  line  passing  through 
the  point  PQ  :  (x0,  y0,  z0)  and  having  the  direction 
cosines  cos  a,  cos  ft,  cos  y.  Let  P  :  (x,  y,  z)  be  an 
arbitrary  point  of  the  line  other  than  P0  and  let 
r  be  the  algebraic  distance  from  P0  to  P,  positive 
if  the  direction  from  P0  to  P  is  that  of  the  line 
and  negative  if  this  direction  is  opposite  to  that 
of  the  line. 

FIG.  7  The  projections  of  P^P,  each  divided  by  r,  are 

equal  respectively  to  the  direction  cosines  of  the 
line,  by  Ch.  XVIII,  §  1,  Th.  2.     Thus 


o  =  Cos  a,         -          =  cos  p,         —     =  cos  y. 
r  r  r 

These  equations  can  be  put  into  the  form 

(1)  x  =  x0  +  r  cos  a,         y  =  y0  +  r  cos  /?,         z  =  z0  +  r  cos  y. 

Equations  (1)  give  the  coordinates  (x,  y,  z)  of  the  point  P  on 
the  given  line  at  the  arbitrary  distance  r  from  P0.  If  r  is 
allowed  to  vary  through  all  values,  positive,  zero,  and  nega- 
tive, P  takes  on  all  positions  on  the  line,  and  always  its 
coordinates  are  given  by  equations  (1).  These  equations,  then, 
represent  the  line.  Since  they  express  the  coordinates  of  the 
point  P  :  (x,  y,  z)  tracing  the  line  in  terms  of  the  auxiliary 
variable,  or  parameter,  r,  we  call  them  a  parametric  representa- 
tion of  the  line. 

If  the  line  is  determined  by  the  point  P0  :  (x0)  y0,  z0)  and  its 
direction  components  I,  m,  n,  we  have,  according  to  (2),  §  3, 
the  following  parametric  representation 

(2)  x  =  x0  +  pl,        y  =  y0  +  pm,,        z  =  zQ  +  Pn. 

The  parameter  p  is  not,  in  general,  equal  to  the  distance  from 
P0  to  P  :  (x,  y,  z),  but  is  merely  proportional  to  this  distance.* 

*  Cf.  footnote,  p.  479. 


THE    STRAIGHT    LINE 


491 


The  Helix.     Given  the  cylinder 

a?  +  y*  =  a2 

with  the  axis  of  z  as  axis.  The  circle  in  the  (#,  y)-plane  on 
which  the  cylinder  is  erected  has  the  parametric  representa- 
tion (Ch.  VII,  §  10)  : 

x  —  a  cos  0,        y  =  a  sin  0,        z  =  0, 

where  0  is  the  angle  which  the  radius,  OP,  to  the  point 
P  :  (x,  y,  0)  makes  with  the  positive  axis  of  x.  On  the  ruling  of 
the  cylinder  through  P  mark  the  point  Pf  at  a  height  above  P* 
equal  to  a  constant  multiple,  IcO,  of  the 
angle  0.  The  coordinates  of  P'  are 

(3)      a;= 


When  0  =  0,  P  and  P1  coincide  in  the 
point  P0  on  the  axis  of  x.  As  0  increases 
from  0  to  2  TT,  the  point  P  traces  the  circle, 
and  the  point  P',  always  directly  above  P, 
traces  a  locus  on  the  cylinder,  encircling  it 
just  once.  When  0  increases  from  2-n-  to 
47r,  P  retraces  the  circle,  whereas  P'  con- 
tinues on  its  rising  path,  encircling  the 
cylinder  a  second  time.  Consequently, 
when  0  increases  through  all  positive  values, 
the  locus  traced  by  P'  encircles  the  cylinder 
infinitely  many  times. 

As   9   decreases    from   zero   through  all 
negative  values,  P',  starting  from  P0,  en- 
circles the  lower  half  of  the  cylinder  infinitely  many  times. 
The  complete  locus  of  P1  is,  then,  an  unbroken  curve  continu- 
ously winding  about  the  cylinder  in  both  directions.     It  is  this 
curve  which  is  represented  parametrically  by  the  equations  (3). 

Since  the  height  of  P'  above  (or  below)  P  is  always  propor- 
tional to  the  angle  0  through  which  the  radius  OP  of  the  circle 
has  turned,  the  curve  (3)  which  P'  traces  is  mounting  on  the 
*  Above  P,  if  0  is  positive  ;  below  P,  if  6  is  negative. 


FIG.  8 


492  ANALYTIC   GEOMETRY 

cylinder  with  a  uniform  steepness.  It  is  ,the  curve  of  the 
thread  of  a  machine  screw  and  is  called  a  circular  screw  or  a 
circular  helix. 

The  Twisted  Cubic.  Consider  the  curve  represented  para- 
metrically  by  the  equations 

(4)  x  =  at,        y  =  bP,        z  =  cP, 

where  t  is  the  parameter,  and  a,  6,  c  a're  constants,  not  zero. 
Like  the  helix,  this  curve  is  a  twisted  curve.  It  is  known  as 
a  twisted  cubic. 

Points  of  Intersection  of  a  Curve  and  a  Surface.  Example  1. 
The  straight  line  of  §  6  can  be  represented  parametrically  by 
setting  each  of  the  members  of  the  continued  equality 


equal  to  a  parameter  t  and  by  solving  the  three  resulting 
equations  for  x,  y,  z: 

(5)  x  =  1  1  —  1,         y  =  t  —  1,         z  =  7  —  %t. 

A  point  of  this  line  lies  in  the  plane  of  §  6, 
2x  +  3y  —  2z  +  l  =  0,  '    • 

if  and  only  if  its  coordinates,  as  given  by  (5),  satisfy  the  equa- 
tion of  the  plane  ;  i.e.  if  and  only  if  t  is  a  solution  of  the 
equation 


or  f«-18  =  0. 

Hence  t  =  4.  But  the  point  t  =  4  of  the  line  (5),  i.e.  the  point 
corresponding  to  the  value  4  for  the  parameter  t,  has  the 
coordinates  : 

a?  =  2  -  1  =  1,         y  =  4  -  1  =  3,         z  =  7  -  1  =  6. 
Hence  the  line  intersects  the  plane  in  the  point  (1,  3,  6). 

Example  2.     Find  the  points  of  intersection  of  the  twisted 
cubic 


THE   STRAIGHT   LINE  493 

(6)  x  =  t,        y  =  t\        z  =  t* 

with  the  plane  2x  +  y  —  z  =  0. 

A  point  of  the  cubic  lies  in  the  given  plane  when  and  only 
when  its  coordinates,  as  given  by  (6),  satisfy  the  equation  of 
the  plane.  Hence  the  solutions  of  the  equation 


for  t  determine  all  the  points  of  intersection. 

One  solution  is  t  =  0  ;  the  others  are  t  =  2,  t  =  —  1.  The 
points  of  the  cubic  (6)  corresponding  to  these  values  of  t  are 
respectively  (0,  0,  0),  (2,  4,  8),  (-  1,  1,  -  1).  Thus  the  cubic 
meets  the  plane  in  these  three  points  and  nrno  further  point. 

The  above  method  may  be  used  to  find  the  points  of  inter- 
section of  any  given  curve  with  a  given  surface,  provided  the 
curve  is  defined  by  a  parametric  representation.  The  sim- 
plicity and  effectiveness  of  the  method  is  one  of  the  advan- 
tages of  representing  a  curve  parametrically. 

.  EXERCISES 

Find  a  parametric  representation  for  each  of  the  following 
straight  lines. 

1.  Through  (2,  —  3,  5)  with  the  direction  cosines  ^,  —  ^,  ^. 

2.  Through   (2,    —3,   5)    with   the   direction   components 
2,  -6,3. 

3.  Through  (ajb  y1}  zj,  (x«,  yz,  z2)  ;  cf.  (1),  §  4. 

4.  3z-4?/-6z  +  7  =  0,  2x-y-2z  +  l  =  Q. 
Suggestion.     First  find  the  direction  components  of  the  line 

and  the  coordinates  of  a  point  on  it. 

The  same  for  the  lines  of  the  following  exercises  of  previous 
paragraphs. 

5.  Ex.  1,  §  3.  7.    Ex.  1,  §  4.  9.    Ex.  1,  §  2. 

6.  Ex.  2,  §  3.  8.   Ex.  2,  §  4.  10.   Ex.  3,  §  2.- 


494  ANALYTIC   GEOMETRY 

Find  equations  for  the  lines  with  the  following  parametric 
representations. 

11.  x  =  3  —  2t,     ?/  =  £  — 4,     z  =  3t  +  2. 

12.  x  =  3t,    y  =  5t,    z=—8t. 

13.  If  from  the  point  (3,  2,  —  6)  one  proceeds  12  units  in 
the  direction  whose  cosines  are  -|,  — »^,  f,  what  are  the  coordi- 
nates of  the  point  reached  ?  Ans,    (11,  —  2,  2). 

14.  Draw  to  scale  the  circular  helix  for  which  a  =  4,  k  =  2. 

15.  Show  that  the  twisted  cubic  (6)  is  the  total  intersection 
of  the   parabolic   cylinder  y  =  x2   with   the   cylinder  z  =  x3. 
Hence  construct  the  cubic. 

16.  Find  a  parametric  representation  of  the  curve 

2/2  =  2x,  z  =  3y3. 

17.  The  same  for  the  curve  of  §  1,  Example  4. 
Suggestion.     Let  x  =  sin  2  6. 

By  the  method  of  this  paragraph  find,  in  each  of  the  follow- 
ing exercises,  the  point  (or  points)  of  intersection  of  the  given 
curve  and  the  given  surface.  * 

18.  The  line  and  plane  of  Ex.  1,  §  6. 

19.  The  line  and  plane  of  Ex.  3,  §  6. 

20.  The  line  and  cylinder  of  Ex.  6,  §  7. 

EXERCISES    ON    CHAPTER    XX 

1.  Find  the  equations  of  the  line  which  passes  through 
the  point  (1,  —2,  3)  and  intersects  the  axis  of  z  at  right 
angles. 

Find  the  coordinates  of  the  points  in  which  the  given 
line  meets  the  coordinate  planes  and  hence  construct  the 
line. 

f    x  +  5y-     «-7  =  0,  x-l-y+2-2-z 

=  U.       3"  ~  3" 


THE   STRAIGHT   LINE  495 

Show  that  the  first  of  the  two  following  lines  intersects  the 
axis  of  z  and  that  the  second  intersects  the  axis  of  x. 


[2x  +  3y-    z  +  2  =  0,  x  +  4  =  2y  +  6  =  3z  +  4 

j    x  _  2y  +  2z-±  =  0.  4  3  2 

6.    What  is  the  condition  that  the  line 
A&  +  B^y  +  CiZ  +  Dl  =  0,        A2x  +  B2y  +  C2z  +  Z>2  =  0, 
where  (71Cf2  =£  0,  meet  the  axis  of  z  ?        ^4ws.  CiZ)2  —  CZD1  =  0. 

Show  that  the  following  lines  are  identical. 


2x  -  2y  +  z  -  1  =  0. 


9.    Find  the  equations  of  the  altitudes  of  the  tetrahedron  of 
Ch.  XIX,  §  6,  Ex.  9. 

10.  Find  the  equation  of  the  plane  which  contains  the  point 
(2,  —  1,  5),  is  perpendicular  to  the  plane  2x  —  ?/  +  3z  =  4,  and 
is  parallel  to  the  line 

5x  +  2y  +  3z  =  0,         4*  +  y  +  2z  -  8  =  0. 

Ans.     3z-9y-5z  +  10  =  0. 

11.  Find  the  equation  of  the  plane  which  passes  through  the 
points  (2,  —  1,  3),  (5,  0,  2)  and  is  parallel  to  the  line 

2z-5  =  l-2/  =  2-3z. 

12.  Find  the  equations  of  the  line  which  contains  the  point 
(4,  2,  —  3),  is  parallel  to  the  plane  x  +  y  +  z  =  0,  and  is  per- 
pendicular to  the  line  whose  equations   are   x  +  2y  —  2  =  5, 
2  =  4.  Ans.     6a-24  =  3?/-6=  -  2z  -6. 

13.  A  line  is  parallel  to  the  plane  2x  —  3y  +  4  =  0.     If  the 
perpendicular  from  the  origin  on  the  line  meets  it  in  the  point 
(2,  5,  —  3),  what  are  the  equations  of  the  line  ? 

14.  Show   that   the   equation   of   the   plane   which    passes 
through  the  point  (oj0,  y0,  z0)  and  is  parallel  to  two  (non-parallel) 


496  ANALYTIC   GEOMETRY 

lines  having  llf  m1}  nt  and  1%,  ra2,  w2  as  direction  components 
can  be  written  in  the  form 


x  — 


15.  Two  lines  with  the  direction  components  lt,  raj,  HI  and 
12,  m-2,  n2  intersect  in  the  point  (a:0,  y0,  z0).     Find  the  equation 
of  the  plane  containing  them. 

16.  Find  the  equations  of  the  line  determined  by  the  point 
(2,  —  1,  0)  and  the  point  of  intersection  of  the  three  planes 


17.  Find  the  equations  of  the  line  through  the  origin  and 
the  point  of  intersection  of  the  plane  and  the  line  whose  equa- 
tions are  x  -}-2y  —  3  z  +  4  =  0  and  3  cc  -f-  1  =  2  —  2y  =  z  +  3. 

18.  Determine  the  equations  of  the  line  which  lies  in  the 
plane  2x  —  y  +  z  —  3  =  0  and  is  perpendicular  to  the  line 


in  the  point  in  which  this  line  meets  the  plane. 

Ans      a-2_y-l_   _  z 
12          19  5 

19.  A  line  through  the  origin  with  the  direction  cosines 
T>   —  f  >  f  intersects  the  plane  3x  +  5y  +  2z  —  6  =  0  in  the 
point  P.     Find  the  length  of  OP.  Ans.     14. 

Suggestion.     Represent  the  line  parametrically. 

20.  A  line  through  the  point  A  :  (3,  —  2,  5)  with  the  direc- 
tion cosines  -f  ,  -|-,  f  meets  the  plane  2x  +  3y  —  z  +  7  =  0  in  the 
point  P.    What  is  the  length  of  AP? 

Loci 

21.  Find  the  locus  of  a  point  which  is  always  equidistant 
from  the  three  points  (2,  0,  3),  (0,  -  2,  1),  (4,  2,  0). 


THE   STRAIGHT   LINE  497 

22.  Determine  the  point  in  the  plane  x  —  y  —  22  =  0  which 
is  equidistant  from  the  three   points    (2,  1,  5),  (4,   —  3,  1), 
(-2, -1,3).  Ans.     (1,1,  i> 

23.  Show  that  the  locus  of  a  point   moving  so  that  it  is 
always  equidistant  from  three  given  non-collinear  points  is  a 
line  perpendicular  to  the  plane  of  the  three  points.     In  what 
point  does  it  intersect  this  plane  ? 

Suggestion.     Choose  the  coordinate  axes  skillfully. 

24.  Find  the  locus  of  a  point  which  is  equidistant  from  the 
points   (2,  3,  0),  (4,  —  1,  2)  and   also   equidistant  from   the 
points  (5,  2,  -  3),  (3,  0,  1). 

25.  The  previous  problem  for  any  four  non-coplanar  points 
PI,  PI  and  P3,  P4.     Show  that  the  locus  is  a  line  which  is 
perpendicular    to   each    of   the   lines   PiP2,    P%Pt   and    goes 
through  the  center  of  the  sphere  determined  by  the  four  points. 

26.  Find  the  locus  of  a  point  which  moves  so  that  the  dif- 
ference of  the  squares  of  its  distances  from  two  given  points  is 
constant. 


CHAPTER   XXI 

THE    PLANE   AND    THE    STRAIGHT  LINE. 
ADVANCED    METHODS 

1.  Linear  Combination  of  Two  Planes.  A  linear  combination 
of  two  planes, 

(1)  Ajx  +  Biy  +  dz  +  A  =  0, 

(2)  A2x  +  B2y  +  C2z  +  A  =  0, 
shall  be  defined  as  any  plane 

(3)  A!  (Atfc  +  B$  +  dz  4-  ^i)  +  \2(A2x  +  B^y  +  dz  +  -^2)  =  0> 

whose  equation  is  obtained  by  multiplying  the  equations  of 
the  two  planes  by  constants,  Al5  A2,  and  adding  the  results. 
The  constants  A1?  A2  can  be  chosen  at  pleasure,  provided  merely 
that  the  coefficients  of  x,  y,  z  in  (3),  namely  \iAt  +  A2v42> 
A^!  +  A2.B2,  X-id  +  A2C2,  do  not  all  vanish.  In  particular,  the 
case  AX  =  A2  =  0  is  thus  excluded. 
For  example,  if  the  given  planes  are 

(4)  x-2y-2z  +  9  =  0, 

(5)  2x-3y-2z  +  S  =  0, 

and  we  multiply  the  equation  of  the  first  by  —  3  and  the  equa- 
tion of  the  second  by  2  and  add,  the  plane  defined  by  the 
resulting  equation, 

(6)  x  +  22-11  =  0, 

is  a  linear  combination  of  the  given  planes. 

If  the  planes  (1)  and  (2)  meet,  the  plane  (3)  passes  through 
their  line  of  intersection.  For,  if  (a;0,  y0  z0,)  is  an  arbitrary 
point  of  this  line,  then 

498 


ADVANCED    METHODS  499 

Bflo  +  CJ.ZQ  +  A  =  0, 


A2x0  +  BO/*  +  (7220  +  A  =  0, 


since  the  point  lies  in  both  the  planes  (1)  and  (2).  If  it  is  also 
to  lie  in  the  plane  (3),  the  equation 

\I(A!XO  +  Btfo  -t-  <7iZ0  +  A)+  A2(^2z0  +  B2y0  +  C2z0  +  A)=  0 

must  be  a  true  equation  ;•  and  this  it  is,  since  the  two  paren- 
theses on  the  left-hand  side  both  vanish  by  virtue  of  equa- 
tions (7).  Thus  the  plane  (3)  contains  the  arbitrary  point 
(XQ,  y0,  z0)  on  the  line  of  intersection  of  the  planes  (1)  and  (2) 
and  hence  contains  the  whole  line,  q.  e.  d. 

We  have  thus  proved  the  theorem  : 

THEOREM  1.  A  linear  combination  of  two  intersecting  planes 
is  a  plane  through  their  line  of  intersection. 

For  example,  the  plane  (6)  passes  through  the  lice  of  inter- 
section of  the  planes  (4)  and  (5). 

If  the  planes  (1)  and  (2)  are  parallel,  it  follows,  by  Ch.  XIX, 
§  7,  Th.  2,  that 

A2  =  pAi,        B2  =  pB1}         Co  =  pCi,  p=£  0. 

The  direction  components  of  the  normals  to  the  plane  (3), 
namely, 

\\AI  +  \2A2)       AI.BI  -\-  A2.B2,       Xjv/i  +  A202, 

become,  then, 

(Ai  +  p\2}  Al}         (A!  +  pAOA,          (A!  -1-  PA2)  C:. 

Now  A!  +  p\z  =£  0,  since  otherwise  the  coefficients  of  x,  y,  z  in 
(3)  would  all  be  zero  ;  that  is,  in  this  case  we  must  exclude, 
according  to  the  definition,  not  only  the  values  A!  =  A2  =  0  but 
also  the  values  of  A!  and  A2  for  which  Ax/A2  =  —  p.  It  follows, 
then,  that  the  plane  (3)  is  parallel  to  or  identical  with  the 
plane  (1).  Thus  we  have  the  theorem  : 

THEOREM  2.  A  linear  combination  of  two  parallel  planes  is 
a  plane  parallel  to  them  or  coincident  with  one  of  them. 

Plane  through  a  Line  and  a  Point.  It  is  now  a  simple  matter 
to  find  the  equation  of  a  plane  determined  by  a  line  and  a  point. 


500  ANALYTIC   GEOMETRY 

For  example,  let  the  line  be  the  line  of  intersection  of  the 
planes  (4)  and  (5)  and  let  the  point  be  (—5,  —  1,  2).  By 
Th.  1,  the  plane 

(8)        A.1(aj-22/-2z  +  9)+A2(2a;-3i/-2z  +  8)=0 

passes  through  the  given  line.     If  it  is  also  to  contain  the 
given  point,  (—5,  —  1,  2),  we  must  have 

A!  (-  5  +  2  -  4  +  9)+  A2(-  10  +  3  -  4  +  8)=  0, 
2A1-3X2  =  0. 

This  equation  determines  the  ratio  Ai/A2.     It  will  be  satisfied 
if,  in  particular,  we  take  AI  =  3  and  A2  =  2.     Then  (8)  becomes 

3 (a;  -  2y  -  2z  +  9)+2'(2a;  -  3y  -  2z  +  8)=  0, 
or  7z-12y- lOz +  43  =  0. 

This  is  the  equation  of  the  required  plane. 

In  the  general  case,  when  the  given  line  is  the  line  common 
to  two  intersecting  planes,  (1)  and  (2),  and  (xi,  y1}  z^  is  the 
given  point,  not  on  the  line,  the  procedure  is  quite  the  same. 
The  plane  (3)  passes  through  the  given  line.  Demanding, 
further,  that  it  contain  the  point  (x1}  yi}  Zj)  leads  to  an  equa- 
tion for  the  determination  of  the  ratio  Ar/A2,  and  any  values 
for  A!  and  A2  which  have  this  ratio  yield,  when  substituted  in 
(3),  the  equation  of  the  required  plane. 

Converses  of  Theorems  1,  2.  Since  every  plane  through  the 
line  of  intersection  of  the  given  planes  (1)  and  (2)  can  be 
thought  of  as  determined  by  this  line  and  a  point  (xh  ylt  z^) 
external  to  it,  we  have  proved  that  every  plane  through  the 
line  of  intersection  of  the  given  planes  is  a  linear  combination 
of  them.  This  is  the  converse  of  Theorem  1. 

The  converse  of  Theorem  2  can  be  proved  in  a  similar 
manner.  The  details  are  left  to  the  student.  Both  converses 
can  be  stated  in  a  single  theorem. 

THEOREM  3.  Any  plane  through  the  line  of  intersection  of 
two  intersecting  planes,  or  parallel  to  two  parallel  planes,  is  a 
linear  combination  of  the  two  planes. 


ADVANCED   METHODS  501 

Projecting  Planes  of  a  Line.     Consider  the  line  L : 

I    x-2y-2z  +  Q  =  0, 
\2x-3y-2z  +  S  =  0, 

in  which  the  planes  (4)  and  (5)  intersect,  and  also  the  plane  (6) : 
x  +  2  z  -  11  =  0. 

This  plane  passes  through  L,  since  it  is  a  linear  combination  of 

the  planes  (4)  and  (5).     In  particular,  it  is  the  plane  through 

L  which  is  perpendicular  to  the  (z,  x)- 

plane,  for  equation  (6)  contains  no  term 

in  y.     It  is,  then,  the  plane  which  projects 

L  on  the  (z,  o;)-plane.     Accordingly,  it  is 

known  as  a  projecting  plane  of  L. 

Equation  (6)  represents,  in  space,  this 
projecting  plane.  Considered  merely  in 
the  (z,  x)-plane,  it  defines  the  line  which 
is  the  actual  projection  of  L  on  the  FIG.  1 

(z,  #)-plane. 

By  combining  equations  (4)  and  (5)  linearly  so  that  the  re- 
sulting equation  contains  no  term  in  x,  e.g.  by  multiplying 
the  first  of  the  equations  by  —  2  and  adding  it  to  the  second, 
we  obtain  the  equation  of  the  plane  which  projects  L  on  the 
(y,  z)-plane,  namely, 

(10)  y  +  2  z  -  10  =  0. 

In  a  similar  manner  the  equation  of  the  plane  which  projects 
L  on  the  (x,  y)-plane  is  found  to  be 

(11)  s-y-l  =  0. 

The  planes  (6),  (10),  and  (11)  are  the  three  projecting  planes 
of  L.  Any  two  of  the  three  projecting  planes  of  a  line  will, 
in  general,*  determine  the  line.  For  example,  the  pair  of 
equations 

(12)  a?  +  2  2  -  11  =  0,        y  +  2z-  10  =  0 

*  Exceptions  occur  when  the  line  is  parallel  to,  or  lies  in,  a  coordinate 
plane  ;  cf .  Ex.  13. 


502  ANALYTIC   GEOMETRY 

is  as  proper  an  analytic  representation  of  L  as  the  pair  (9), 
and  much  more  simple. 

Equations  (6),  (10),  and  (11)  are  equivalent  to  the  continued 
equality 

(13)  x  -  11  =  y  -  10  =  -  2  z, 

and  conversely.  In  other  words,  the  representation  of  a  line 
by  means  of  its  projecting  planes  is  essentially  the  same  as 
the  representation  of  it  by  means  of  a  continued  equality  of 
the  usual  form.  The  three  equations  which  result  from 
equating  the  members  of  the  equality  are  the  equations  of 
the  projecting  planes. 

Furthermore,  we  now  have  a  method  of  finding,  from  the 
representation  of  a  line  as  the  intersection  of  two  planes,  a 
representation  of  it  by  a  continued  equality.  Thus,  in  the 
case  of  L,  we  passed  from  the  equations  (9)  to  the  continued 
equality  (13). 

EXERCISES 

Find  the  equation  of  the  plane  determined  by  the  given 
line  and  the  given  point. 

Line  Point 

j2*-32,  +  4z-2  =  0,  0 

- 


, 

2z  =  S,  (3,  -  1,  2). 

3.   3  x  -  5  y  =  6,  2  x  +  3  z  =  9,  (4,  3,  -  5). 

4  a+i_y-i_z-3  n      1  2^ 

5  -2  "       4  (1»~1>2)- 

5.  Find  the  equation  of   the  plane   containing  the  line  of 
Ex.  1  and  having  the  intercept  2  on  the  axis  of  y. 

6.  Find  the  equation  of  the  plane  passing  through  the  line 
of  Ex.  2  and  having  equal  intercepts,  not  zero,  on  the  axes  of 
x  and  y.  Ans.    13  x  +  13  y  +  22  z  +  8  =  0. 


ADVANCED  METHODS  503 

7.  What  is  the  equation  of  the  plane  which  contains  the  line 
of  Ex.  4  and  is  perpendicular  to  the  plane  3x  —  y  +  4  2  =  0  ? 

Ans.   <ix-\-8y  —  z  —  1  =  0. 

8.  Find  the  equations  of  the  line  which  is  the  projection  of 
the  line  of  Ex.  3  on  the  plane  2x  +  y  —  3  z  +  5  =  0. 

9.  Find  the  equations  of  the  projecting  planes  of  the  line 
of  Ex.  1  and  from  them  determine  a  continued  equality  which 
represents  the  line. 

10.  The  preceding  exercise  for  the  line  of  Ex.  2. 

11.  The  line  of  Ex.  3  is  denned  by  two  of  its  projecting 
planes.     What  is  the  equation  of  the  third  ? 

12.  What  are  the  equations  of  the  projecting  planes  of  the 
line  of  Ex.  4  ? 

13.  A  line  which  is  not  parallel  to  or  in  a  coordinate  plane 
has  three  projecting  planes,  which  are  distinct ;  a  line  parallel 
to  or  in  one  coordinate  plane  has  three  projecting  planes,  just 
two  of  which  are  identical ;   a  line  parallel  to  or  in  two  coordi- 
nate planes  —  i.e.  parallel  to  or  coincident  with  an  axis  —  has 
but  two  projecting  planes,  which  are  distinct.     Consequently 
there  are  always  at  least  two  distinct  projecting  planes  of  a  line 
and  the  line  is  determined  by  them.     Prove  these  statements. 

14.  A   line   not   parallel  to  or  in  the  (x,  y)-plane  can  be 
represented  by  equations  of  the  form 

x  =  az  -f-  b,        y  =  cz  +  d ; 

a  line  parallel  to  or  in  the  (x,  y)-plane  can  have  its  equations 
put  into  the  form 

y  =  ax  -j-  b,        z  =  c, 

unless  it  is  parallel  to  or  coincident  with  the  y-axis ;  in  this 
case,  its  equations  can  be  written  as 

x  =  a,         z  =  b. 
Prove  these  statements. 

15.  Prove  that  the  plane  determined  by  the  point  (x2,  y2,  z2) 
and  the  line  through  the  point  (x1}  y1}  z})  with  the  direction 


504  ANALYTIC  GEOMETRY 

components  Z,  m,  n  can  have  its  equation  written  in  the  form 
x  —  X!       y  -yi       z  - 


m 


=  0. 


Suggestion.  Determine  the  direction  components  of  the 
normals  to  the  plane. 

2.  Three  Planes  through  'a  Line.  Three  Points  on  a  Line.' 
By  means  of  the  results  of  the  preceding  paragraph  we  can 
prove  the  following  theorem. 

THEOREM  1.  Three  planes  pass  through  a  line  or  are  paral- 
lel, when  and  only  when  any  one  of  them  is  a  linear  combina- 
tion of  the  other  two. 

If  the  three  planes  pass  through  a  line,  or  are  parallel,  any 
one  of  them  passes  through  the  line  of  intersection  of  the 
other  two,  or  is  parallel  to  the  other  two.  Consequently,  by 
Th.  3,  §  1,  this  plane  is  a  linear  combination  of  the  other  two. 

Conversely,  if  a  particular  one  of  the  planes  is  a  linear  com- 
bination of  the  other  two,  it  goes  through  the  line  of  intersec- 
tion of  these  two,  if  they  intersect,  or  is  parallel  to  them,  if 
they  are  parallel  (Ths.  1,  2,  §  1).  It  follows  then,  further- 
more, by  the  first  part  of  the  proof,  that  any  one  of  the  three 
planes  is  a  linear  combination  of  the  other  two,  q.  e.  d. 

For  example,  the  three  planes, 

3a;_2y+    z+    6  =  0, 

(1)  2a?  +  6y-3«-    2  =  0, 

4x-9y  +  5z  +  14  =  0, 

pass  through  a  line,  inasmuch  as  the  equation  of  the  third  can 
be  obtained  by  multiplying  that  of  the  first  by  2,  that  of  the 
second  by  —  1,  and  by  adding  the  results. 

Three  Points  on  a  Line.  The  three  points  P1 :  (xlt  ylf  Zi), 
P2 :  (#2,  y2>  22)>  PS  '•  (xs>  ys)  zs)  lie  on  a  line  if  and  only  if  the 

*  It  is  assumed,  here  and  in  §  4,  that  the  given  planes,  or  the  given 
points,  are  distinct. 


ADVANCED   METHODS  505 

direction   components   of   PiP3   are   proportional  to  those  of 


*!>,     y,  —  yi  =  p(yz  -  yi),     z3-z,=  P(z2  -  z$. 
These  equations  can  be  rewritten  as  • 


(2)  2/3=(l-p) 

z3  =  (1  —  p)  «i  +  p  «2- 

They  then  say  that  the  coordinates  of  P3  are  a  linear  combina- 
tion of  the  coordinates  of  PI  and  P2  with  constants  of  combina- 
tion, 1  —  p  and  p,  ivhose  sum  is  unity.  Since  any  one  of  the 
points  might  have  been  called  P3,  this  result  can  be  stated 
more  generally. 

THEOREM  2.  Three  points  lie  on  a  line  when  and  only  when 
the  coordinates  of  any  one  of  them  can  be  expressed  as  a  linear 
combination  of  those  of  the  other  two,  with  constants  of  combina- 
tion whose  sum  is  unity. 

This  theorem  is  of  importance  because  of  the  analogy  be- 
tween it  and  Theorem  1,  and  because  of  its  theoretical  value 
in  later  work.  It  has  not  the  practical  value  of  Theorem  1, 
since  testing  three  points  for  collinearity  can  be  more  easily 
done  directly. 

Thus,  if  the  three  points  are  (2,  —  1,  5),  (4,  2,  6),  (-2,  -7,  3), 
the  direction  components  of  PjP2  and  PiP3  are,  respectively, 
2,  3,  1  and  —  4,  —  6,  —  2.  Since  these  triples  are  propor- 
tional, the  three  points  lie  on  a  line. 

EXERCISES 

What  can  you  say  of  the  three  planes  in  each  of  the  follow- 


1.  2x  —  y  —  z  =  2,  3x 

2.  x  +  3y  —  z  =  l,  3x-5y  +  7z  =  3,  3x  +  2y  +  2z  =  3. 

3.  6x~3y  +  9z  =  2,  2x-y  +  3z  =  Q,  -±x+2y-  6z=3. 

Are  the  three  given  points  collinear  ? 


506  ANALYTIC   GEOMETRY 

4.  (5,  3,  4),  (1,5,  10),  (11,0,  -5). 

5.  (-  13,  12,  -  15),  (-  5,  6,  -  11),  (7,  -  3,  -  5). 

6.  (2,  -3,8),  (5,4,7),  (8,10,6). 

7.  Determine  k  so  that  the  three  planes 

kx—  3y  +  z  =  2,     3x  +  2y  +  4z  =  l,    x  —  Sy  —  2z  =  3 
will  pass  through  a  line. 

8.  Determine  k  so  that  the  three  points  (2,  3,  A;),  (5,  5,  1), 
(  —  1,  1,  9)  will  be  collinear. 

3.  Line  in  a  Plane.  From  Theorem  1  of  the  preceding  para- 
graph follows  immediately  the  theorem : 

THEOREM.  A  line  lies  in  a  plane,  if  and  only  if  the  plane  is  a 
linear  combination  of  any  two  planes  which  determine  the  line. 

For  example,  the  line  of  intersection  of  the  first  two  of  the 
planes  (1),  §  2  lies  in  the  third  plane,  since  the  third  plane 
was  shown  to  be  a  linear  combination  of  the  first  two. 

A  second  method  of  testing  whether  or  not  a  given  line  lies 
in  a  given  plane  presents  itself  if  the  line  is  represented  para- 
metrically.  The  line  through  the  point  (2,  —  1,  3)  with  the 
direction  components  3,  2,  —  4  has  the  parametric  representa- 
tion (Ch.  XX,  §  8,  (2))  : 

x  =  3t  +  2,        y  =  2t-l,        z  =  -4:t  +  3. 

It  will  lie  in  the  plane 

2x  +  5y  +  4z-  11  =  0, 

if  and  only  if  the  coordinates  of  every  one  of  its  points  satisfy 
the  equation  of  the  plane,  i.e.  if  and  only  if 

2(3 1  +  2)  +  5(2 1  -  1)  +  4(-  4 1  +  3)  -  11  =  0 

is  a  true  equation  for  all  values  of  t.     But  the  equation  re- 
duces to 

0  .  t  +  0  =  0, 

and  so  is  satisfied  by  all  values  of  t.     Consequently,  the  line 
lies  in  the  plane. 


ADVANCED   METHODS 


507 


It  is  clear  that  this  method  can  also  be  applied  to  test 
whether  or  not  a  given  curve,  represented  parametrically,  lies 
on  a  given  surface. 

EXERCISES 

In  each  of  the  following  exercises  determine  whether  or  not 
the  given  line  lies  in  the  given  plane.  Apply  both  methods. 


Line 

x-l_y+_2_z_+3, 
~T~        -3~     -1  ' 


2.  2x-3  =  l-5y  =  3 

3.  ±x-l=-3y  =  l- 


Plane 

v_z_2-0. 

4x  +  5y  -  3z  -  7  =  0. 
2x-y  +  3z  +  5  =  Q. 


4.  Find  the  conditions  under  which  the  line  through  the 
point  (o;0,  y0)  20)  with  the  direction  components  I,  m,  n  will  lie 
in  the  plane  Ax  +  By  +  Cz  +  D  =  0. 

Ans.     Al  +  Bm  +  Cn  =  0,  Ax0  +  By0  +  Cz0  +  D  =  0. 

5.  Does  the  twisted  cubic  x  =  t ,  y  =  t1,  z  =  2fi  lie  on  the 
surface  2  x3  —  z  =  0  ? 

6.  Show  that  the  curve 

x  =  a  sin2 1,        y  =  a  sin  t  cos  £,         2  =  a  cos  t 
lies  on  the  sphere  ce2  +  y2  +  z2  =  a2. 

4.  Four  Points  in  a  Plane.  Four  Planes  through  a  Point. 
Given  the  four  points  Plf  P2,  P3>  P4,  with  the  coordinates 

(«i>  2/i>  *i)>    (^2,  2/2,  2*2)  >    (^3,  2/3,  2s),    (^4,  2/4,  «0-      To   determine 
when  they  lie  in  a  plane. 

Unless  all  four  points  lie  on  a  line,  there  will  be  three  of 
them  which  determine  a  plane.  Let  these  three  be  P2,  P3,  P4. 
The  equation  of  the  plane  through  them  is,  by  Ch.  XIX,  §  6, 

1 


x     y 

«2       2/2 
*3       2/3 


=  0. 


ANALYTIC   GEOMETRY 


The  four  points  will  be  coplanar  if  and  only  if  Pl :  (xl}  yi} 
lies  in  this  plane,  i.e.  if  and  only  if 


(1) 


2/1 

2/2 
2/3 
2/4 


=  0. 


If  the  four  points  do  lie  on  a  line,  they  are  certainly  coplanar. 
On  the  other  hand,  equation  (1)  is  satisfied  in  this  case  also. 
For,  equations  (2),  §  2,  must  hold  for  some  value  of  p,  since 
the  points  P1}  P2,  Ps  are  collinear.  Consequently,  if  in  the 
determinant  in  (1)  we  subtract  from  the  third  row  the  first 
row  multiplied  by  1  —  p  and  the  second  row  multiplied  by  p, 
the  new  third  row  will  consist  exclusively  of  zeros,*  and  hence 
the  determinant  will  vanish. 

We  have  proved,  then,  the  following  theorem. 

THEOREM  1.  The  four  points  (a?i,yi,«i),  (afc,  y2,  z2),  (x3,  y3,  z3), 
(c»4,  2/4,  z4)  lie  in  a  plane  if  and  only  if  the  determinant  in  (1) 
vanishes. 


Four  Planes  through  a  Point.     Let  the  four  planes  be 


(2) 


A3x 


B2y 

B3y 


C3z  +  D3  =  0, 
<74z  +  Z>4  =  0. 

Form  the  determinant  of  the  coefficients  in  these  equations, 
namely,  \AlB2C3T)/^\  or,  more  simply,  \ABCD  \.     In  this  de- 
terminant let  AU  A2,  A3,  A4  be  the  minors  of  the  elements  D1} 
D2,  Z>3,  D4  ;  for  example,  A4  =  |  A^B^C^  |-t 
We  first  prove  the  following  Theorem  : 
THEOREM  2.     The  normals  to  the  planes  (2)  will  all  be  parallel 
to  a  plane  if  and  only  if 

Aj  =  A2  =  A3  =  A4  =  0. 

*  The  fourth  element  in  the  new  third  row  is  1  —  (1  —  p)  —  p  =  0. 
t  Throughout  this  paragraph  we  denote  determinants  by  writing  the 
elements  of  their  principal  diagonals  between  two  parallel  bars. 


ADVANCED   METHODS  509 

For,  if  the  normals  to  the  four  planes  are  parallel  to  a  plane, 
the  normals  of  any  three  are  parallel  to  this  plane,  and  hence 
the  determinant  of  their  direction  components  vanishes,  by  Ch. 
XVIII,  §  6.  But  this  determinant  is  one  of  the  four  determi- 
nants A  in  question.  Hence  all  four  determinants  A  vanish. 

Conversely,  if  all  four  determinants  A  vanish,  the  normals 
of  each  set  of  three  planes  are  parallel  to  a  plane,  and  this 
plane  can  be  taken  as  the  same  plane  in  all  four  cases.* 
Hence  the  normals  to  all  four  planes  are  parallel  to  it. 

Suppose,  now,  that  the  planes  have  one  and  only  one  point 
in  common.  Then  the  four  determinants  A  are  not  all  zero. 
For,  if  they  were  all  zero,  the  normals  to  the  four  planes  would 
all  be  parallel  to  a  plane.  Consequently,  either  the  four 
planes  would  be  all  parallel  or  all  the  lines  of  intersection 
obtained  by  taking  them  in  pairs  would  be  parallel  or  identical 
(of.  Ch.  XIX,  §  10),  and  hence  the  planes  would  have  either 
no  point  in  common  or  a  -whole  line  of  points  in  common. 
But  this  contradicts  the  hypothesis  that  they  meet  in  just  one 
point.  At  least  one  of  the  determinants  A,  then,  does  not 
vanish.  Let  us  assume,  say,  that  A4  =  AiBzC3  \  is  not  zero. 
Then  the  first  three  planes  meet  in  a  single  point  (Ch.  XIX, 
§  10),  whose  coordinates,  found  by  Cramer's  rule,  are 


or 


,f  = 

A4 


Since  this  point  lies  in  the  fourth  plane,  we  must  have 


*  This  is  obvious  if  the  four  planes  are  parallel.  In  the  contrary  case, 
when  at  least  two  of  the  planes,  say  the  first  two,  are  not  parallel,  the  state- 
ment is  substantiated  as  follows.  The  normals  to  the  first  two  planes  and 
the  third  are  parallel  to  a  plane  M\,  and  the  normals  to  the  first  two  planes 
and  the  fourth  are  parallel  to  a  plane  M2.  But  M\  and  M2,  since  they 
are  both  parallel  to  the  normals  of  the  first  two  planes  are,  in  any  case, 
parallel  to  each  other  and  hence  can  always  be  taken  as  the  same  plane. 


510  ANALYTIC  GEOMETRY 

or,  since  the  expression  on  the  left  is  the  development  of  the 
determinant  |  ABCD  \  by  the  minors  of  the  last  row, 

(4)  \ABCD\=0. 

Conversely,  if  |  ABCD  \  =  0  and  not  all  four  of  the  determi- 
nants A  are  zero,  the  planes  (2)  meet  in  a  single  point.  For, 
we  can  assume  that  A4  =£  0.  Then  the  first  three  planes  meet 
in  a  single  point  (3)  and  this  point  lies  in  the  fourth  plane, 
since  by  hypothesis  (4)  holds. 

Thus  we  have  proved  the  theorem  : 

THEOREM  3.  The  four  planes  (2)  meet  in  a  single  point  if  and 
only  if  the  determinant  of  their  coefficients  vanishes  and  not  all 
four  minors  A1?  A2,  A3,  A4  are  zero.* 

If  the  normals  to  the  planes  (2)  are  all  parallel  to  a 
plane,  the  determinant  \ABCD\  obviously  vanishes,  for  then 
A!  =  A2  =  A3  =  A4  =  0  and  the  expansion  of  |  ABCD  \  by  the 
minors  of  the  fourth  column,  namely: 


has  the  value  zero. 

Conversely,  if  |  ABCD  [  vanishes  by  virtue  of  the  vanishing 
of  A!,  A2,  A3,  A4,  the  normals  of  the  planes  (2)  are,  by  Theorem 
2,  all  parallel  to  a  plane. 

Consequently,  we  can  combine  Theorems  2  and  3  in  the  more 
general,  though  less  useful,  theorem  : 

THEOREM  4.  The  four  planes  (2)  meet  in  a  single  point  or 
their  normals  are  all  parallel  to  a  plane,  if  and  only  if  the  deter- 
minant of  their  coefficients  vanishes. 

Finally,  we  enumerate  the  cases  which  can  occur  when  the 
normals  to  the  four  planes  are  parallel  to  a  plane.  First,  the 

*  Stated  algebraically  this  theorem  reads  :  The  four  equations  (2) 
are  compatible  and  have,  moreover,  a  unique  solution,  if  and  only  if 
|  ABCD\  =  0  and  AI,  A2,  A3,  A4  are  not  all  zero.  This  theorem  includes 
the  theorem  of  Ch.  XVI,  §  9,  Ex.  8,  and  also  its  converse.  It  is  to  be 
noted  that  it  was  geometric  considerations  which  led  us  here  to  a  proof 
which  covered  the  converse  as  well  as  the  theorem. 


ADVANCED   METHODS  511 

planes  can  be  all  parallel  ;  this  case  can  easily  be  detected  by 
inspection.  Secondly,  the  planes  can  all  go  through  a  line; 
the  test  of  §  3  reveals  this  case.  Finally,  whatever  lines  of  in- 
tersection the  planes  have,  when  taken  in  pairs,  are  all  paral- 
lel ;  this  case  will  make  itself  known  by  exclusion  of  the  others. 

Example.     Consider  the  four  planes 

x+    y  +  3z  —    6  =  0, 

±x  +  2y-5z  +    8  =  0, 

Sy-  Iz  +  22  =  0, 

2x-3y  +    z-    7  =  0. 

Here  |  ABCD  \  =  0  and  A4  =£  0,  as  can  easily  be  verified. 
Hence  the  four  planes  meet  in  a  single  point.  Let  the  student 
show  further  that  the  last  three  planes  pass  through  a  line, 
which  is  intersected  by  the  first  plane  in  the  point  in  ques- 
tion. 

EXERCISES 

Do  the  four  given  points  lie  in  a  plane  ?  If  so,  do  three, 
or  do  all  four,  lie  on  a  line? 

1.  (2,  3,  1),  (1,  5,  2),  (-  3,  4,  -  1),  (-  2,  2,  -  2). 

2.  (2,  5,  3),  (0,  2,  -  3),  (1,  3,  7),  (-  1,  -  1,  15). 

3.  (1,  2,  -  1),  (3,  1,  2),  (-  1,  3,  -  4),  (7,  -  1,  8). 

4.  (0,  2,  1),  (1,  0,  2),  (-  1,  -  1,  1),  (4,  2,  3). 

5.  For  what  values   of  k  will  the  four  points  (k,  —  5,  6), 
(4,  _  4,  k),  (5,  1,  2),  (2,  0,  7)  lie  in  a  plane  ? 

What  can  you  say  of  the  relative  positions  of  the  four  given 
planes  ? 

6.  The  planes  of  Ch.  XVI,  §  9,  Ex.  9. 

7.  The  planes  of  Ch.  XVI,  §  9,  Ex.  10. 


2x  +  ±y-3z-  1  =  0,  x—2y  +  3z-2  =  0, 

5x-    y-    z  +  2  =  0,  '   3x+    y  +  2z  +  3  =  0, 

=  0.  5x  +  4+    z  -  5  =  0. 


512  ANALYTIC   GEOMETRY 

10.   For  what  value  of  k  will  the  four  planes, 

kx+   y  —  z  —  6  =  0,           x  —  y  +    z  =  0, 
x  +  ty  +  z  —  3  =  0,         2x  +  y  +  4:Z  —  1  =  0, 

go  through  a  point? 

5.  Two  Intersecting  Lines.     Given  the  two  distinct  lines 

(  Ajas  +  B$  +  C,z  +  A  =  0,      f  AJx  +  BSy  +  Ci'z  +  A'  =  0, 
1  A2x  +  JS-ji/  +  C2z  +  D»  =  0;     1  ^'a;  +  JBj'y  +  Cz'z  +  Dz'  =  0. 

The  two  lines  intersect  in  a  point,  when  and  only  when  the 
four  planes  which  in  pairs  determine  them  meet  in  a  single 
point.  The  condition  for  this  is  given  in  Theorem  3  of  the 
preceding  paragraph. 

The  two  lines  are  parallel  if  and  only  if  the  normals  to  the 
four  planes  are  all  parallel  to  a  plane.  Theorem  2  of  §  4  tells 
when  this  occurs. 

These  results  can  be  combined  in  the  general  theorem : 

THEOREM  1.  Two  lines  intersect  or  are  parallel  when  and 
only  tuhen  the  determinant  of  the  coefficients  in  the  equations  of 
the  four  planes  which  in  pairs  determine  the  two  lines  vanishes. 

The  simplest  way  to  decide  in  any  case  whether  the  two 
lines  intersect  or  are  parallel  is  to  compute  the  direction  com- 
ponents of  the  lines  and  compare  them. 

The  above  proof  assumes  tacitly  that  the  four  planes  in 
question  are  distinct ;  otherwise  the  theorems  of  §  4  could 
not  be  applied.  Theorem  1  still  holds,  however,  in  the  ex- 
ceptional case  when  one  of  the  planes  determining  one  line 
is  identical  with  one  of  the  planes  determining  the  other  line. 
For,  the  two  lines  lie,  then,  in  a  plane  and  hence  intersect  or 
are  parallel ;  on  the  other  hand,  the  determinant  in  question 
contains  two  rows  which  are  proportional  and  hence  it  vanishes. 

Lines  Given  by  Continued  Equalities.  Let  the  first  line  be 
determined  by  the  point  P1 :  (xl}  yit  Zj)  and  the  direction  com- 
ponents llf  mi,  «!  and  the  second  by  the  point  P2 :  (a^,  yz,  z%) 
and  the  direction  components  1%,  m2,  n2. 


ADVANCED   METHODS 


513 


The  lines  are  parallel  if  and  only  if  11}  mi}  Wj  are  proportional 
to  12)  wi2,  w2. 

If  the  lines  are  not  parallel,  there  is  a  unique  plane  which 
contains  the  first  line  and  is  parallel  to  or  contains  the  second. 
A  normal  to  this  plane  is  perpendicular  to  both  lines  and 
hence  has  the  direction  components  |  m^  |,  |  n^z  \,  \  ZjW^  |  ;  cf. 
Ch.  XVIII,  §  5.  The  equation  of  the  plane  is,  then, 

(1)  |  ra^a  |  (as  —  »i)+  |  nj,z  \  (y  -  y^+  \  ^2  \  («  —  «i)  =  0, 
or 

x-xl 

(2)  raj          HI 

If  this  plane  contains  the  point  P2  it  wiH  contain  the  entire 
second  line,  and  conversely.  Consequently,  the  two  non-parallel 
lines  intersect  if  and  only  if 


=  0. 


(3) 


m1 


=  0. 


Equation  of  the  Plane  Determined  by  Two  Intersecting  or 
Parallel  Lines.  If  two  given  lines  intersect  or  are  parallel,  the 
plane  in  which  they  lie  can  be  determined  by  one  of  the  lines, 
say  the  first,  and  a  point  of  the  second  which  does  not  lie  on 
the  first.  Its  equation,  then,  can  be  found  by  the  method  of 
§  1.  This  method,  though  always  applicable,  is  designed 
primarily  for  the  case  when  at  least  one  of  the  lines  is  given 
as  the  intersection  of  two  planes. 

If  both  lines  are  represented  by  continued  equalities,  the 
plane  which  they  determine  has  (2)  as  its  equation,  in  case  the 
lines  intersect.  If  the  lines  are  parallel,  a  similar  equation 
for  their  plane  can  be  found  ;  cf.  §  1,  Ex.  15. 

EXERCISES 

Show  that  the  given  lines  intersect  or  are  parallel.  In  each 
case  find  the  equation  of  the  plane  which  they  determine. 


514 


1. 


ANALYTIC   GEOMETRY 

2x  —  y  —  z  =  0, 


Ans.   The  plane  is  5  a;  —  4  y  —  2  z  -f  1  =  0. 

1     x  —  y  —  z  =  0;  1  3cc-f  3y— z +  4  =  0. 

z-6 


y-3=y-2=z-l 
2       "  -5          3 


— 1 
__ 


2 
4a- 


1=0, 


6.   Distance  of  a  Point  from  a  Line.    Distance  between  Two 
Lines.     Let  it  be  required  to  find  the  distance  D  of  the  point 
P2 :  fa,  yz,  z2)  from  the  line  L  which  passes 
through  the  point  PI  :  (x1}  yit  Zj)  and  has  the 
rL    direction  cosines  cos  a,  cos  /?,  cos  y. 
It  is  clear  from  the  figure  that 

D  =  PiP2  sin  0, 

FIG.  2  where  0  is  the  angle  which  the  line  P^PZ  makes 

with  L. 
By  Ex.  16  at  the  end  of  Ch.  XVIII, 

(1)  sin2  6  =  |  piv2 12  +  |  Vl\2 12  +  |  A.i/12 12, 

where  A.J,  ^  vi  and  A2,  p*,  v2  are  the  direction  cosines  of 
and  L. 

Now, 


Hence 


Similar  values  are  found  for  j  viA.2 1  a>nd  |  Aj^  |. 


A,-- 

2  -  *1                            2/2  —  .Vl 

Z2  —  Zj 

P  P'    '                             P  P     ' 

1/1  ~  P  P  ' 

A2  =  cos  a,             /i2  =  cos  /?, 

V2  =  COS  y. 

!••.    = 
"  ~ 

y*  —  y\.    «2  —  ^i 

1 

COS  )8          COS  y 

2/2  —  2/1      Z2  —  zl 
COS  (3          COS  y 

AA 

ADVANCED   METHODS 


515 


Substituting  these  values  in  (1),  multiplying  the  resulting 
equation  through  by  P]_P{-  and  extracting  the  square  root  of 
both  sides,  we  obtain,  as  the  final  result  : 


COS  /?        COS  y 


COS  y 


COS  a 


cos  a     cos 


Distance   between    Two  Skew  Lines.     Let  the  line  through 
the  point  P{  :  (xlt  yi}  2^)  with  the  direction  components  Z1}  m^  nt 
and  the  line  through  the  point 
P2  :  (#2>  2/2>  Z2)  with  the  direction 
components    1%,   7^2,   713    be    two 
skew  lines,  r'.e.  two  lines  which 
neither  intersect  nor  are  paral- 
lei.     To  find  the  distance  D  be- 
tween them  measured  along  their  common  perpendicular. 

The  plane  through  the  first  line  parallel  to  the  second  has 
the  equation  (1),  §  5,  namely, 


FIG.  3 


(x  - 


-  «=  0. 


The  required  distance  D  is  the  uniform  distance  of  the  sec- 
ond line  from  this  plane,  or  it  is  the  distance  of  the  point 
P2  :  (#2,  2/2>  z2)  from  this  plane.  Thus, 


2)  —  ± 


|  (ya  — 


or 


where  A  is  the  determinant  in  fprmula  (3),  §  5,  and  where 
that  sign  is  to  be  chosen  which  will  make  the  right-hand  side 
positive. 

Distance  between  Two  Parallel  Lines.  The  distance  between 
two  parallel  lines  can  be  found  as  the  distance  of  a  point 
on  one  of  the  lines  from  the  other  line. 


516  ANALYTIC   GEOMETRY 

EXERCISES 

Find  the  distance  of  the  given  point  from  the  given  line. 
Point  Line 

1.  (2,3,4),  1=^  =  1=:^=*.    Ans.  f  VlOl  =  4.31. 

2.  (0,0,0),  *= 


/  —  O 

3.    (-1,2,  -3),      3a  +  l=4 


5.  (3,  1,  -  1), 

Find  the  distance  between  the  two  given  lines. 

6.  The  lines  of  Exs.  ^  2.  7.    The  lines  of  Exs.  2,  3. 

Ans.    M  V89  =  1.59. 

o  y 

8.    The  lines  of  Exs.  1,  3.  9.    The  lines  of  Exs.  2,  4. 

10.    The  lines  of  Exs.  3,  4.  11.    The  lines  of  Exs.  4,  5. 

12.  A  cube  has  edges  of  length  a.  Find  the  distance  be- 
tween a  diagonal  and  an  edge  skew  to  it.  Ans.  ^  V2  a. 

7.   Area  of    a  Triangle.    Volume  of   a  Tetrahedron.     We 

first  prove  the  following  theorem. 

THEOREM  1.  If  a  region  of  area  A  in  a  plane  M  is  projected 
on  a  second  plane  M',  the  area  of  the  projected  region  equals 
A  cos  6,  where  0  is  the  acute  angle  between  M  and  M'.* 

In  the  case  of  a  rectangle  whose  sides  are  respectively 
parallel  and  perpendicular  to  the  line  of  intersection  of  M 
and  M' ,  the  proof  is  immediate.  For,  when  the  rectangle  is 
projected  on  M',  one  dimension  remains  the  same  and  the 
other  is  multiplied  by  cos  B. 

*  The  theorem  is  trivial  if  M  is  parallel  or  perpendicular  to  M' ;  we 
exclude  these  cases. 


ADVANCED   METHODS 


517 


The  area  A  of  an  arbitrary  region  in-  M  is  the  limit  ap- 
proached by  the  sum  B  of  the  areas  of  rectangles,  of  the  type 
just  described,  which  are  inscribed 
in  the  region  : 

A  =  lim  B. 

If  A'  is  the  area  of  the  projected 
region  and  B'  is  the  sum  of  the 
areas  of  the  projected  rectangles, 
evidently 

A'  =  lim  B'. 

Since  the  area  of  each  projected 
rectangle   is  cos  0  times  the   area  of  the  original  rectangle, 

B'  =  B  cos  e. 

Hence  A'  =  lim  B  cos  6  =  cos  0  lim  B 

or  A'  =  A  cos  6,  q.  e.  d. 

Let  the  areas  of  the  projections  of  the  given  region  on  the 
coordinate  planes  be  denoted  by  Ayz,  A,x)  Axy,  and  let  the 
normals  to  M  have  the  direction  angles  a,  (3,  y.  By  Th.  1,* 

Av.  =  |  A  cos  a  |  ,  Atx=\Acosp\,  Azy  =  |  A  cos  y  \  . 

Hence 

(1)  A^AJ  +  AJ  +  AJ. 

Thus  we  have  proved  the  theorem  : 

THEOREM  2.  The  sum  of  the  squares  of  the  areas  of  the  pro- 
jections of  a  region  on  the  three  coordinate  planes  equals  the 
square  of  the  area  of  the  region. 

Area  of  a  Triangle.  It  is  now  easy  to  write  down  a 
formula  for  the  area  A  of  the  triangle  whose  vertices  are  at 
the  points  P^  :  (x,,  yl}  2^,  P2  :  (x2,  y.2,  z2),  Pz  :  (x3,  y3,  z3).  For, 
the  areas  of  the  projections  of  the  triangle  on  the  coordinate 
planes  are,  by  Ex.  18  at  the  end  of  Ch.  XVI, 


*  The  absolute  value  signs  are  necessary  since  a,  /3,  y  are  not  neces- 
sarily acute  angles. 


518 


ANALYTIC   GEOMETRY 


where  |  i/fa  1 1 ,  for  example,  is  the  determinant  whose  three 
columns  are  ylf  y2,  y3 ;  zlt  z2,  z3 ;  1,  1,  1.     Hence,  by  (1), 


(2)  A  = 

Volume  of  a  Tetrahedron.     Let  the  above  triangle  be   the 
base  of  the  tetrahedron  and  let  the  fourth  vertex  be  at  the 
point  P0 :  (x0,  y0,  z0).     The  volume   V 
of    the    tetrahedron    is    known    from 
Solid   Geometry   to   be  equal   to   one 
third  the  area  A  of  the  base  times  the 
.y    length  D  of  the  altitude : 
(3)  V=\AD. 

The  equation  of  the  plane  of  the 
base,  the  plane  of  Pl}  P2,  P3,  is  given 
in  determinant  form  in  Ch.  XIX,  §  6. 

This   equation,  when   the  determinant   is  developed   by  the 

minors  of  the  first  row,  becomes 

=  0. 


FIQ.  5 


The  distance  of  the  point  P0  :  (»0,  yQ,  z0)  from  this  plane,  i.e. 
the  length  D  of  the  altitude  of  the  tetrahedron,  is,  by  Ch. 
XIX,  §  9, 

(4)      D  =  ±  I  yiZ*  •*•  I  x°~ 


Substituting  in  (3)  the  values  of  A  and  D  as  given  by  (2) 
and  (4),  and  at  the  same  time  writing  the  numerator  in  (4)  in 
determinant  form,  we  obtain,  as  the  value  of  V, 


2/2 
2/3 


28 


where  that  sign  is  to  be  taken  which  yields  a  positive  result. 


ADVANCED   METHODS  519 

EXERCISES 

Find  the  areas  of  the  following  triangles. 

1.  With  vertices  at  (2,  - 1,  3),  (4,  3,  -  2),  (3,  0,  -  1). 

Ans.   iVl34  =  5.79. 

2.  With  vertices  at  (0,  0,  0),  (xly  y^  zj,  (a^,  y2,  z2). 

3.  Cut  from  the  plane  2  x  —  3  y  +  4  z  —  12  =  0by  the  coor- 
dinate planes. 

4.  With  vertices  at.  (a,  0,  0),  (0,  b,  0),  (0,  0,  c). 

Find  the  volumes  of  the  following  tetrahedra. 

5.  That  of  Ch.  XIX,  §  6,  Ex.  9. 

6.  That  of  Ch.  XIX,  §  10,  Ex.  5. 

7.  Included  between  the  plane  2x  —  3  ?/  +  4  z  =  12  and  the 
coordinate  planes. 

8.  Included  between  the  coordinate  planes  and  the  plane 
with  intercepts  a,  b,  c  on  the  axes. 

9.  With  vertices  at  (0,  0,  0),  (xlt  ylt  Zi),  (a^,  yz,  Zj),  (a^,  ?/3,  Zs). 

EXERCISES  ON  CHAPTER  XXI 

1.  Find  a  parametric  representation  of  the  line 

x—  y  —  z  +  1  =  0,        2x  —  y  +  z  —  8  =  0. 

2.  What  are  the  equations  of  the  projecting  planes  of  an 
arbitrary  line  passing  through  the  origin  ? 

3.  Find  the  equation  of  the  plane  which  contains  the  line  of 
Ex.  1  and  is  parallel  to  the  line  2x  —  3  =  y  —  3  =  2z—  1. 

4.  Find  the  equation  of  the  plane  which  is  perpendicular  to 
the  plane  2x  +  5y  —  3z  —2=0  and  meets  it  in  the  line  in 
which  it  intersects  the  (x,  y)-plane. 

Ans.    6a  +  15y  +  29z-  6  =  0. 

5.  Do  Ex.  16  at  the  end  of  Ch.  XX  without  finding  the 
coordinates  of  the  point  of  intersection  of  the  three  given  planes. 

Suggestion.     Find  two  planes  through  the  given  point,  each 
containing  the  line  of  intersection  of  a  pair  of  given  planes. 


520  ANALYTIC   GEOMETRY 

6.  Do  Ex.  17  at  the  end  of  Ch.  XX  without  finding  the 
coordinates  of  the  point  of  intersection  of  the  given  line  and  the 
given  plane. 

7.  Find  the  equations  of  the  line  which  contains  the  point 
(2,  0,  —  1)  and  intersects  each  of  the  lines 


-2z=2;  3x+±y-  z  +  1  =  0. 

8.  Find  the  equations  of  the  line  which  intersects  each  of  the 
lines  given  in  Ex.  7  and  is  parallel  to  the  line  4  —  6x  =  y  +  $  —  2z. 

9.  A  plane  intersects  the  (x,  ?/)-plane  in  the  line  whose  equa- 
tion in  the  (x,  ?/)-plane  is  2  x  +  3  y  =  12.     If  the  plane  cuts  from 
the  first  octant  a  tetrahedron  whose  volume  is  12,  find  its 
equation.  Ans.   2x  +  3y  +  4:Z  —  12  =  0. 

10.  There  are  two  planes  which  contain  the  line 

x  +  2y  +  z  +  ~L=Q,         2x+y-z-7=Q 

and  make  angles  of  30°  with  the  plane  x  —  z  +  2  =  0.     Find 
their  equations.     Ans.  x  —  y  —  2z—8  =  0,     2x-\-y  —  z  —  7  =  Q. 

11.  Find  the  equation?  of  the  planes  which  contain  the  line 
given  in  Ex.  10  and  are   V2  units  distant  from  the  point 

(2,2,  -3). 

12.  The  planes  through  the  edges  of  a  trihedral  angle  perpen- 
dicular to  the  opposite  faces  pass  through  a  line.     Prove  this 
theorem  in  the  case  that  the  faces  lie  in  the  planes 


13.  Prove  the  theorem  of  Ex..  12  in  the  general  case. 

THE  EQUATIONS  A.W  -f  pv  =  0,  uv  =  0  * 

14.  THEOREM.     If  u  =  0,  v  =  0  are  the  equations  of  two  sur- 
faces, the  equation  \u+  /ru=0,  A/A^tO,  represents  in  general]  a 


*Cf.  Ch.  IX,  §§3,  4. 

t  In  particular,  it  may  represent  a  curve  or  a  point  ;  cf  .  footnotes,  pp. 
445,  167. 


ADVANCED   METHODS  521 

surface  which  contains  the  total  intersection  of  the  two  surfaces, 
if  they  intersect,  and  has  no  other  point  in  common,  ivith  either  of 
them.  If  the  given  surfaces  do  not  intersect,  the  equation  repre- 
sents in  general  *  a  surface  not  meeting  either  of  them  or  it  has  no 
locus.  Prove  this  theorem. 

15.  Find  the  equation  of  the   sphere  which   contains   the 

circle 

&  _|_  f  _|_  22  _  4  _  o,         2-5  =  0 

and  passes  through  the  point  (3,  0,  2). 

Ans.   x2  +  y*  +  z2  +  3  z  =  19. 

16.  Prove  that  the  curve  of  intersection  of  the  cylinders 
xz  +  yz  =  4,  y2  +  z2  =  5  lies  on  the  surface  x2  —  z2  -f  1  =  0  and  is 
the  total  intersection  of  this  surface  with  each  cylinder. 

17.  THEOREM.     If  u  =  0,  v  =  0,  w  =  0  are  the  equations  of 
three   planes    which    meet    in    a    single    point,    the    equation 
\u  +  fj,v  +  vw  =  0  represents  a  plane  through  this  point.     Con- 
versely, every  plane  which  contains  this  point  is  a  linear  combina- 
tion of  the  three  given  planes.     Prove  this  theorem. 

18.  Find  the  equation  of  the  plane  which  is  determined  by 
the  points  (0,  0,  0),  (1,  2,  0)  and  the  point  of  intersection  of  the 
three  planes 

2x  +  y  +  3z  +  l  =  Q,   x  +  y-4z  +  2  =  Q,  2x-2y  +  5z  —  3  =  0. 

Ans.   2x  —  y  —  6z  =  0. 

19.  THEOREM.     The  equation  uv  =  0  represents  those  points 
and  only  those  points  which  lie  on  the  surfaces  u  =  0  and  v  =  0. 
Prove  this  theorem. 

20.  What  do  the  following  equations  represent? 

(a)  a;2  —  y2  =  0  ;     (c)   x2  —  xy  —  xz  +  yz  =  0 ; 
(6)  x*  -  y*  =  0  ;  -  (d)  xy  -  xz  -  2y +  2z  =  0. 

*  In  particular,  it  may  represent  a  curve  or  a  point ;  cf .  footnotes,  pp. 
445,  168. 


522  ANALYTIC   GEOMETRY 

BISECTORS  OP  THE  ANGLES  BETWEEN  Two  PLANES  * 

21.  What  is  the  locus  of  the  inequality 

2X-y  +  2z  —  4>0? 

22.  There  are  four  regions  lying  between  the  planes 

2x-y  +  2z-±  =  Q,         8a;  +  4y  +  z-8  =  0. 

Find  the  pairs  of  simultaneous  inequalities  representing  these 
regions,  specifying  the  region  which  each  pair  represents. 

23.  Find  the  equations  of  the  planes  bisecting  the  angles  be- 
tween the  two  planes  of  Ex.  22. 

Ans.   2x  +  7y  —  52  +  4  =  0,     14ar  +  y  +  7z  -  20  =  0. 

24.  The  same  for  the  following  pairs  of  planes 

x-y  +  z-2  =  0,  3x-Gy  +  2z-±  =  0, 

)  x  +  y-z  +  3  =  0-,         w  6x  +  2y-$z-5  =  0. 

25.  Find  the  equation  of  that  bisector  of  the  angle  between 
the  two  planes  of  Ex.  24  (6)  which  passes  through  the  region  be- 
tween the  two  planes  which  contains  the  origin.  -.    • 

26.  The  planes  which  bisect  the  dihedral   angles  of  any 
trihedral  angle  meet  in  a  line.  .  Prove  this  theorem  when  the 
faces  lie  in  the  planes 

x  +  y  +  z  —  1  =  0,         x  —  y+z—  1  =  0,         2x  +  y  —  z  +  l  =  0. 

27.  Prove  the  theorem  of  Ex.  26  in  the  general  case. 
Suggestion.     Cf.  Exs.  28,  29  at  the  end  of  Ch.  XIII  and 

Ex.  28  at  the  end  of  Ch.  XIX. 

28.  Show  that  by  a  proper  choice  of  axes  two  arbitrarily 
chosen  skew  lines,  LI  and  L2,  can  have  their  equations  written  as 

x  =  c,     z  —  my\        x  =  —  c,     z  =  —my;        cm  =/=0. 

29.  Prove  that,  if  the  line  L^  of  Ex.  28  is  taken  as  the  2-axis, 
the  x-  and  y-axes  can  be  so  chosen  'that  Lz  has  the  equations, 
x  =  c}  z  =  my,  where  c  =£  0. 

»Cf.  Ch.  XIII,  §§6,  7,  8. 


CHAPTER   XXII 

SPHERES,  CYLINDERS,  CONES.      SURFACES  OF 
REVOLUTION 

1.  Equation  of  the  Sphere.  The  equation  of  the  sphere 
whose  center  is  at  the  origin  and  whose  radius  is  p  is,  accord- 
ing to  Ch.  XIX,  §  1, 

(1)  z2+y2  +  22  =  /02. 

It  can  be  shown  in  a  similar  manner  that,  if  the  center  is 
at  the  point  (a,  /?,  y)  and  the  radius  is  p,  the  equation  of  the 
sphere  is 

(2)  (aj  -  «)t  +(y  -  fl*+(z  -  y)2  =  f, 

Thus  the  sphere  whose  center  is  at  the  point  (2,  —  3,  4) 
and  whose  radius  is  6  has  the  equation 

(x  -  2)2  +(y  +  3)2  +(Z  -  4)2  =  36, 
or  a;2_j_ 


EXERCISES 

Find  the  equations  of  the  following  spheres  and  reduce  the 
results  to  their  simplest  form. 

1.  Center  at  (3,  1,  2)  ;  radius,  5. 

Ans.   o;2  +  r/2  +  z2_  6z-  2y  —  4«—  11  =  0. 

2.  Center  at  (—  2,  3,  —  6)  ;  radius,  7. 

3.  Center  at  (4,  0,  0)  ;  radius,  4. 

4.  Center  at  (0,  —  5,  0)  ;  radius,  2. 

5.  Center  at  (0,  —  4,  3)  ;  radius,  5. 

6.  Center  at  (f  ,  —  ^,  0)  ;  radius,  1. 

523 


524  ANALYTIC   GEOMETRY 

7.  Center  at  (£,  —  f  ,  £)  ;  radius,  f  . 

8.  Center  at  (0,  0,  a)  ;  radius,  a. 

9.  Center  at  (a,  0,  a)  ;  radius,  aV2. 
10.  Center  at  (a,  a,  a)  ;  radius,  aV3. 

2.  General  Form  of  the  Equation.  The  equation  of  a  sphere 
can  always  be  written  in  the  form 

(1)  x2  +  f-  +  z2  +  Ax  +  By  +  Cz  +  D  =  0, 

as  is  seen  by  expanding  equation  (2),  §  1. 

Let  us  investigate  whether,  conversely,  every  equation  of 
the  form  (1)  represents  a  sphere. 

Consider,  first,  the  particular  equation 

(2)  x2  +  yz  +  z2_2x  +  6y  +  4:Z-35  =  0. 

If  we  complete  the  square  of  the  terms  in  x,  and  do  the  same 
for  the  terms  in  y  and  in  z,  the  equation  becomes 

(3)  (x  -  I)2  +(y  +  3)2  +(«  +  2)2  =  35  +  1  +  9  +  4  =  49. 

This  equation  is  of  the  form  (2),  §  1,  where  a  =  1,  /?  =  —  3, 
y  =  —  2,  p  =  7,  and  hence  represents  a  sphere  whose  center 
is  at  the  point  (1,  —  3,  —  2)  and  whose  radius  is  7. 

If  the  constant  term,  —  35,  in  (2)  is  replaced  by  14,  the 
right-hand  member  of  (3)  becomes  —  14  +  1  +  9  +  4  =  0.  In 
this  case,  then,  we  have 


The  point  (1,  —  3,  —  2)  has  coordinates  satisfying  this  equa- 
tion. For  the  coordinates  (x,  y,  z)  of  any  other  point  at  least 
one  of  the  parentheses  is  not  zero  and  the  left-hand  side  of  the 
equation  is  positive.  Consequently,  the  equation  represents 
the  single  point  (1,  —  3,  —  2)  or,  if  we  define  a  null  sphere  (a 
sphere  of  zero  radius)  as  a  point,  it  represents  a  null  sphere. 
If  the  constant  term,  —35,  in  (2)  is  replaced  by  15,  the 
equation  becomes 

(x-  1)2  +(y  +  3)2  +(2  +  2)2  =  -  1. 


SPHERES   AND    OTHER   SURFACES  525 

Since  the  left-hand  member  of  this  equation  can  never  be 
negative,  no  matter  what  values  are  assigned  to  x,  y,  z,  the 
equation  represents  no  point  whatever  in  space. 

These  three  examples  indicate  what  to  expect  of  the  general 
equation  (1).  On  completing  the  squares  for  the  pairs  of  terms 
in  x,  y,  and  z,  respectively,  in  (1),  the  equation  takes  on  the 
form  (2),  §  1,  where 

A          R_     B  C 

a—  ~TT>      P—  ~"9»      y-  ~n> 

(4) 

z  =  A*  +  &+C*-4:D 

4 

Hence,  we  have  the  following 

THEOREM.  Equation  (1)  represents  a  sphere,  a  single  point, 
or  no  point  whatever,  according  as  the  quantity 


is  positive,  zero,  or  negative.  In  case  it  represents  a  sphere,  the 
coordinates  of  the  center  and  the  square  of  the  radius  are  given 
by  formulas  (4). 

Consider,  more  generally,  the  equation 
(5)  a  (x2  +  y*  +  z2)+bx  +  cy  +  dz  +  e  =  Q. 

If  a  =  0,  but  b,  c,  and  d  are  not  all  zero,  the  equation  repre- 
sents a  plane. 

If  a  =£  0,  the  equation  can  be  divided  through  by  a,  and  it 
then  becomes 

aj2  +  ya  +  aB  +  &a;  +  -3/  +  -«+-  =  0. 
a        a        a        a 

This  equation  is  of  the  form  (1)  and  hence  the  foregoing  con- 
siderations apply  to  it. 

EXERCISES      . 

Determine  what  the  following  equations  represent.  Apply 
in  each  case  the  method  of  completing  the  square.  Do  not 
merely  substitute  numerical  values  in  formulas  (4). 

1.    z2  +  2/2  +  z2  +  4a;-  Qy  —  2z  +  5  =  0. 

Ans.    A  sphere,  radius  3,  with  center  at  (—2,  3,  1). 


526 


ANALYTIC   GEOMETRY 


3. 
4. 
5. 
6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 
14. 


Ans.   The  point  (3,  -  4,  -  2). 
.   No  point. 


+ 


—  4  =  0. 

?/ 
z2  =  2  ax. 


4z  +  6  =  0. 
-8z-  5  =  0. 


=  0. 


3.  Sphere  through  Four  Points.  Through  four  points  which 
do  not  lie  in  a  plane  there  passes  a  single  sphere.  If  the 
points  are  (x1}  ylt  zj,  (a^>,  yz,  z2),  (a;3,  y3,  z3),  (a;4,  j/4,  z4),  the  equa- 
tion of  the  sphere,  in  determinant  form,  is 

x      y      z      1 
xl     yl     Zj     1 

X2       ?/2       Z2       1 

x3     7/3     z3     1 

^4       2/4       Z4       1 

To  prove  this,  develop  the  determinant  by  the  minors  of 
the  elements  of  the  first  row.  The  equation  takes  on  the 
form  (5),  §  2.  The  coefficient  of  a;2  +  y2  +  z2  is  the  determinant 
\xil/2z3 1 1>  and  is  different  from  zero,  since  the  four  points  do 
not  lie  in  a  plane  (Ch.  XXI,  §  4).  Consequently,  the  equa- 
tion represents  a  sphere,  a  point,  or  no  point  whatever. 

But  the  coordinates  of  the  four  given  points  satisfy  the 
equation,  since  the  substitution  for  x,  y,  z  of  the  coordinates 


SPHERES  AND   OTHER   SURFACES  527 

of  any  one  of  these  points  makes  the  first  row  of  the  deter- 
minant identical  with  a  later  row,  so  that  the  determinant 
vanishes.  Therefore  the  equation  actually  represents  a  sphere 
and  this  sphere  is  the  one  through  the  four  given  points. 

EXERCISES 

Find  the  equations  of  the  spheres  through  the  following 
sets  of  four  points. 

1.  (1,  0,  0),  (0,  1,  0),  (0,  0,  1),  the  origin. 

Ans.  x2  +  y2  +  z*  —  x  —  y  —  z  =  Q. 

2.  (1,  1,  1),  (-  1,  1,  1),  (1,  -1,  1),  (1,  1,  -  1). 

3.  (2,  3,  1),  (5,  -  1,  2),  (4,  3,  -  1),  (2,  5,  3). 

4.  The  vertices  of  the  tetrahedron  of  Ch.  XIX,  §  6,  Ex.  9. 

5.  The  vertices  of  the  tetrahedron  formed  by  the  coordinate 
planes  and  the  plane  2x  —  3  ?/  +  4  z  —  12  =  0. 

6.  The  vertices  of  the  tetrahedron  of  Ch.  XIX,  §  10,  Ex.  5. 

7.  When  will  five  points,  no  four  of  which  are  coplanar,  lie 
on  a  sphere  ? 

Do  the  five  given  points  lie  on  a  sphere  ? 

8.  (0,  0,  0),  (-  1,  0,  -  1),  (3,  1,  0),  (2,  4,  -  4),  (3,  3,  -  4). 

9.  (0,  2,  3),  (4,  1,  0),  (-  4,  5,  0),  (1,  5,  -  1),  (4,  2,  -  5). 

4.  Tangent  Plane  to  a  Sphere.  Let  P  be  a  point  of  a  sphere 
and  let  the  radius  to  P  be  drawn.  The  plane  through  P  per- 
pendicular to  the  radius  is  the  tangent  plane  to  the  sphere  at  P. 

If  the  sphere  is 

(1)  X*  +  ?/2  +  22  =  p2 

and  P  has  the  coordinates  (xlt  ylt  Zj),  the  radius  to  P  has  the 
direction  components  £Cj,  ylf  zt.  The  tangent  plane  at  P  is  the 
plane  through  (xl}  y1}  z^,  whose  normals  have  these  direction 
components.  Consequently,  its  equation  is 

«i  G»  -  »0  +yi(y-  yj  +  «i  («  -  «0  =  0, 
or  »!*  +  yfl  +  ZjZ  =  »i2  +  2/i2  4-  «i2. 


528  ANALYTIC   GEOMETRY 

Since  the  point  (x1}  yi}  z^  lies  on  the  sphere  (1), 


and  hence  the  equation  of  the  tangent  plane,  in  final  form,  is 

(2)  X&  +  y$  +  z&  =  p2. 

In  a  similar  manner  the  equation  of  the  tangent  plane  to 
the  sphere 

(3)  (a!_a)2  +  (2,_/j)2  +  (2_y)2  =  p2 

at  the  point  (o^,  yl}  z^)  of  the  sphere  can  be  shown  to  be 

(4)  (Xl  -  a)(x  -  a)  +  (y,  -  ft)(y  -  fi  +  fa  -  y)(«-y)  =  p2. 
The  use  of  (4)  to  find  the  equation  of  the  tangent  plane  to 

a  sphere  whose  equation  is  in  the  form 

(5)  a2  +  y*  +  z2  +  Ax  +  By  +  Cz  +  D  =  0 

involves  the  reduction  of  the  equation  of  the  sphere  to  the 
form  (3).     Thus,  if  the  sphere  is 

(6)  o;2  +  ?/2  +  z2-2a;  +  62/  +  4z-35  =  0, 
the  equation  must  first  be  rewritten  as 

(x-  I)2  +  (y  +  3)2  +  (z  +  2)2  =  49. 

The  equation  of  the  tangent  plane  at  the  point  (3,  —  6,  4),  for 
example,  is  then,  according  to  (4), 

(3  -  l)(a?  -  !)  +  (-  6  +  3)(y  +  3)  +  (4  +  2)(z  +  2)  =49, 
or 

(7)  2«-32/-t-6z-48  =  0. 

The  coordinates  of  the  center  (a,  ft,  y)  and  the  square  of  the 
radius,  p2,  of  a  sphere  whose  equation  is  in  the  form  (5)  are 
given  by  formulas  (4),  §  2.  If  these  values  for  a,  ft,  y,  p2  are 
substituted  in  (4)  and  the  equation  obtained  is  simplified,  the 
result  is 


(8) 

This  is  the  equation  of   the   tangent   plane   at  the   point 
(#!>  yi>  zi)  to  a  sphere  whose  equation  is  in  the  form  (5).     By 


SPHERES  AND   OTHER  SURFACES  529 

means  of  it  the  equation  of  the  tangent  plane  to  (6)  at  the 
point  (3,  —  6,  4)  can  be  written  down  directly.  We  have, 
namely, 

3X  -  Qy  +  4z  -(x  +  3)  +  3(*/  _  6)  +  2(z  +  4)-  35  =  0, 

and  this  reduces  to  the  equation  (7)  obtained  by  the  indirect 
method. 

EXERCISES 

Find  the  equation  of  the  tangent  plane  to  each  of  the  follow- 
ing spheres  at  the  given  point. 

1.  *2  +  y2  +  z2  =  9at(2,  _2,  -1). 

2.  x*  +  f  +  z2  =  49  at  (3,  -  6,  2). 

3.  ,(*  -  I)2  +  (y  ~  2)2  +  (z  +  3)2  =  81  at  (2,  6,  5). 

4.  a?+(y  +  6)»  +  (2-4)*  =  9at(l,  -3,  2). 

6.   a?2  +  y2  +  z2  —  2 x  —  ±y  +  4 z  =  0  at  the  origin. 

6.  z2  +  y2  +  z2-6a;-h4?/  +  10z-ll  =  0at  (1,1,1). 

7.  Find  the  volume  of  the  tetrahedron  cut  from  the  first 
octant  by  the  tangent  plane  at  (1,  2,  3)  to  the  sphere 

2a2  +  2t/2  +  2z2  +  2x  -  3y  -  4z  -  12  =  0. 

8.  The  coordinates  of  one  of  the  points  of  intersection  of  the 
plane  2x  —  y  —  2  =  0  with  the  sphere  of  Ex.  1  are  (2,  2,  1). 
Find  the  angle  between  the  plane  and  the  sphere. 

<t  Ana.   72°  39'. 

9.  Find  the  angle  which  the  line  x  =  y  =  z  makes  with  the 
sphere  of  Ex.  5.  Ans.    11°  6'. 

5.  The  Circle.     A  plane  intersects  a  sphere  in  a  circle,  is 
tangent  to  it,  or  fails  to  meet  it,  according  as  the  distance  D 
of  the  center  of  the  sphere  from  the  plane  is  less  than,  equal 
to,  or  greater  than,  the  radius  p  of  the  sphere. 

In  other  words,  the  equations  of  a  sphere  and  a  plane,  con- 
sidered simultaneously,  represent  a  circle,  a  point,  or  no  point 
whatever,  according  as  D  <  p,  D  =  p,  or  D  >  p. 


530  ANALYTIC   GEOMETRY 

> 
Consider,  for  example,  the  sphere  and  the  plane 


The  center  of  the  sphere  is  at  the  point  (1,  —  3,  —  2)  and  its 
radius  is  p  =  1  ;  cf.  §  2.  The  distance  of  the  center  from  the 
plane  is 


Vl  +  4  +  4  3 

Consequently,  the  plane  meets  the  sphere  in  a  circle,  and  equa- 
tions (1),  considered  simultaneously,  are  the  equations  of  the 
circle. 

It  is  readily  seen  that  pz  —  D2,  i.e.  72  —  62  =  13,  is  the  square 
of  the  radius  of  the  circle.  Hence  the  radius  of  the  circle  is 
Vl3.  The  center  of  the  circle  is  the  point  of  intersection  of 
the  plane  and  the  line  through  the  center  of  the  sphere  perpen- 
dicular to  the  plane.  Its  coordinates  are  thus  found  to  be 
(-1,1,2). 

Radical  Plane  of  Two  Spheres.     Given  the  two  spheres 

a?  +  y*  +  z*  +  2  x  -  2  z  -  1  =  0, 


Z2  +  x  _+_  4y  _  iOz  -  9  =  0. 

Subtract  the  equation  of  the  second  from  that  of  the  first. 
The  resulting  equation, 

(3)  ^-42/4-82  +  2  =  0, 

• 
represents  a  plane.     This  plane  is  known  as  the  radical  plane 

of  the  two  spheres.  Since  its  equation  is  a  linear  combination 
of  the  two  spheres,  we  conclude,  by  Ex.  14  at  the  .end  of 
Ch.  XXI,  the  following  : 

If  the  spheres  intersect,  the  radical  plane  is  the  plane  of 
their  common  circle  ;  if  the  spheres  are  tangent,  it  is  their 
tangent  plane  at  the  point  of  tangency  ;  and  if  the  spheres 
fail  to  meet,*  the  radical  plane  intersects  neither  of  them. 

*  Two  concentric  spheres  have  no  radical  plane  ;  this  is  the  only  ex- 
ceptional case. 


SPHERES  AND   OTHER  SURFACES  531 

Conversely,  the  spheres  intersect  in  a  circle,  are  tangent,  or 
fail  to  meet,  according  as  their  radical  plane  intersects  one  of 
them  in  a  circle,  is  tangent  to  it,  or  fails  to  meet  it.  Thus 
the  question  of  the  relationship  between  two  spheres  is  re- 
duced to  that  of  the  relationship  between  a  plane  and  a  sphere, 
and  this  we  have  already  discussed. 

The  center  of  the  first  of  the  spheres  (2)  is  at  the  point 
(—  1,  0,  1)  and  its  radius  is  3.  The  distance  of  the  center 
from  the  plane  (3)  is  found  to  be  1.  The  radical  plane  and 
the  first  sphere  intersect,  then,  in  a  circle,  and  consequently 
this  is  true  of  the  two  spheres. 

Equations  (2),  considered  simultaneously,  are  a  pair  of  equa- 
tions of  the  circle.  A  simpler  pair  consists  in  one  of  the 
equations  (2)  and  the  equation  (3). 

EXERCISES 

In  each  of  the  following  exercises,  determine  what  the  given 
equations  represent.  If  they  represent  a  circle,  find  its  center 
and  radius  ;  if  they  represent  a  point,  find  its  coordinates. 

1.  x2  +  y1  +  z2  —  25  =  0,  z  =  4. 

2.  xz  +  yz  +  zz  -Qx  —  4  ?/  =  0,  2x  +  y  +  2z  —  1=0. 

3.  z2+?/2  +  z2-4:»-2?/  +  6z-2  =  0, 

4.  x2  +  y*  +  z2  —  2x  =  0,  8a?  —  y  —  3=0. 


Find  the  radical  plane  of  the  spheres  given  in  the  following 
exercises.  If  the  spheres  intersect  in  a  circle,  find  its  center 
and  radius.  If  the  spheres  are  tangent,  find  the  coordinates 
of  the  point  of  contact. 

5.   xz  +  y2  +  &  =  13,  a2  +  f  +  z2  +  3x  -  4  =  0. 

Jz2  +  /+22-2a;  —  4=y  —  Sz  -4  =  0, 
6>     \a?  +  y*  +  z*  +  2y-5z-5  =  Q. 
7.   32+2/2+z2-2  =  0,  2x2+2y2+2z*+3x  —  4:y  +  z  +  2  =  0. 

Find  the  equation  of  the  sphere  determined  by  the  given 
circle  and  the  given  point  ;  cf.  Ex.  15  at  the  end  of  Ch.  XXI. 


532  ANALYTIC   GEOMETRY 

8.  The  circle  of  Ex.  1  and  the  origin. 

9.  The  circle  of  Ex.  2  and  the  point  (1,  1,  1). 
10.    The  circle  of  Ex.  5  and  the  point  (1,  —  2,  3). 

6.  Cylinders.  Given  a  plane  curve,  not  a  straight  line,  and 
through  each  point  of  the  curve  draw  an  indefinite  straight 
line  perpendicular  to  the  plane.  The  surface  generated  by 
these  lines  is  called  a  cylinder.  The  lines  are  its  rulings,  or 
generators,  and  the  given  curve  its  directrix. 

We  shall  consider  here  only  cylinders  whose  rulings  are 
parallel  to  a  coordinate  axis. 

If  the  rulings  of  a  cylinder  are  parallel,  for  example,  to  the 
axis  of  z,  the  equation  of  the  cylinder  does  not  contain  z. 
For,  the  directrix  can  be  thought  of  as  lying  in  the  (x,  y)- 
plane,  and  its  equation  in  this  plane  will  represent  in  space 
the  cylinder,  inasmuch  as  the  points  whose  coordinates 
satisfy  the  equation  are  those  points  and  only  those  points 
which  lie  on  the  directrix,  or  directly  above  or  below  it,  i.e. 
which  lie  on  the  cylinder.  But  this  equation  does  not  con- 
tain z,  q.  e.  d. 

Conversely,  a  curved  surface  represented  by  an  equation  in 
which  z  does  not  appear  is  a  cylinder  whose  rulings  are 
parallel  to  the  axis  of  z.  For,  the  equation  defines  in  the 
(x>  y)-plane  a  curve,  and  in  space  it  represents  those  points 
and  only  those  points  which  lie  on,  or  directly  above,  or 
below,  this  curve,  i.e.  which  lie  on  the  cylinder  erected  on 
the  curve. 

We  have,  then,  the  following  theorem :  A  curved  surface  is 
a  cylinder  with  rulings  parallel  to  a  coordinate  axis  when  and 
only  when  its  equation  does  not  contain  the  variable  correspond- 
ing to  that  axis. 

Quadric  Cylinders.  A  cylinder  whose  directrix  is  a  conic  is 
known  as  a  quadric  cylinder.  In  particular,  it  is  called  elliptic 
(or  circular),  hyperbolic,  or  parabolic,  according  as  the  directrix 
is  an  ellipse  (or  circle),  a  hyperbola,  or  a  parabola. 


SPHERES  AND  OTHER  SURFACES 


533 


Figure  1  shows  the  quadric  cylinders  of  the  three  types, 
whose  equations  are 


The  two  hyperbolic  cylinders  represented  by  the  second 
equation  are  known  as  conjugate  hyperbolic  cylinders,  and  the 
planes 


shown  also  in  Fig.  1,  as  their  common  asymptotic  planes. 

The  elliptic  cylinder,  or  either  hyperbolic  cylinder,  of  Fig.  1 
is  symmetric  in  each  point  of  the  axis  of  z.     That  is,  every 


FIG.  1 


quadric  cylinder  whose  directrix  is  a  central  conic  is  symmetric 
in  each  point  of  the  line  drawn  through  the  center  of  the  conic 
parallel  to  the  rulings.  This  line  is  called  the  axis  of  the 
cylinder. 

Sections  of  Quadric  Cylinders.  The  curve  in  which  a  quadric 
cylinder  is  met  by  a  plane,  M,  which  is  not  parallel  to  the  rul- 
ings, we  shall  call  a  (plane)  section  of  the  cylinder. 

THEOREM  1.  A  section  of  a  quadric  cylinder  is  a  conic  of  the 
same  type  as  the  directrix. 

We  give  the  proof  in  the  case  in  which  the  directrix,  D,  is  a 


534 


ANALYTIC   GEOMETRY 


central  conic.  Let  M  intersect  the  plane  Kof  D  in  the  line  L.* 
As  coordinate  axes  in  K  take  Ox  parallel  to  L  and  Oy  perpen- 
dicular to  L,  as  shown.  The 
equation  of  D,  referred  to  these 
axes,  is  of  the  form  (Ch.  XII, 
§3): 

(3)  Ax*  +  Bxy  +  Cy2  +  F'=0. 
Draw  in  M  the  rectangular 
axes,  O'x',  O'y', whose  projections 
on  K  are  respectively  Ox,  Oy. 
Let  P1 :  (x1,  y')  be  an  arbitrary 
point  of  M  and  let  P :  (x,  y)  be 
its  projection  on  K.  Then 

(4)  x  =  x',         y  =  y'  cos  e, 

where  0  is  the  acute  angle  between  M  and  K. 

Since  D  is  the  curve  in  K  into  which  the  section,  S,  of  the 
cylinder  by  M  projects,  the  equation  of  S  is  obtained  from 
equation  (3)  of  Z)  by  substituting  for  x  and  y  in  (3)  their  values 
as  given  by  (4).  Thus  the  equation  of  S  is 

(5) 


FIG.  2 


Ax'2  +  Bx'y'  cos  0  +  Cy'2  cos2  6  +  F*  =  0. 


This  equation  represents  a  conic  and,  furthermore,  a  conic  of 
the  same  type  as  D,  since  the  discriminant  of  the  quadratic 
terms : 

B2  cos2  0  —  4  AC  cos2  6  =  (B*  —  4:  AC)  cos2  0 

is  of  the  same  sign  as  the '  discriminant,  B1  —  4  AC,  of  the 
quadratic  terms  in  (3),  q.  e.  d. 

It  is  clear  that  (5)  is  independent  of  the  height  00'  at  which 
M  cuts  the  axis  of  the  cylinder.  In  otlier  words,  the  sections  by 
two  parallel  planes  are  congruent  conies. 

Suppose,  now,  that  B2  —  4  AC  >  0  and  that  to  F'  is  given  in 
turn  the  values  1,  —  1,  0.  Then  (3)  represents  in  turn  a  hyper- 

*  If  M  is  parallel  to  A',  the  section  by  M  is  congruent  to  the  directrix  ; 
in  this  case,  then,  no  further  proof  is  required. 


SPHERES  AND   OTHER  SURFACES  535 

bola,  the  conjugate  hyperbola,  and  the  common  asymptotic 
lines  ;  but  this  is  true,  also,  of  (5).  We  have,  then,  the  follow- 
ing result. 

THEOREM  2.  The  sections  of  tico  conjugate  hyperbolic  cylinders 
by  a  plane  M  are  two  conjugate  hyperbolas  whose  common  asymp- 
totes are  the  lines  in  which  M  cuts  the  common  asymptotic  planes 
of  the  cylinders. 

Returning  to  the  general  case,  we  assume  that  there  is  given 
a  second  cylinder  with  vertical  rulings,  whose  directrix,  D,  is 
similar  and  similarly  placed  to  D,  or,  if  D  is  a  hyperbola,  is 
similar  and  similarly  placed  to  D  or  to  the  conjugate  of  D. 
The  equation  of  2),  as  a  curve  in  K,  can  be  written,  according 
to  Ex.  40,  p.  260,  in  the  form : 

(6)  As?  +  Bxy  +  Cy2  +  Dx  +  Ey  +  F  =  0. 

The  equation  of  the  section  S  of  the  second  cylinder  by  the 
plane  M  is,  then, 

(7)  Ad*  +  Bx'y1  cos  0  +  Cy'2  cos2  0  +  Dx'  +  Ey1  cos  6  +  F  =  0. 

Since  equations  (5)  and  (7)  fulfill  the  conditions  of  Ex.  40, 
p.  260,  it  follows  that  S  and  ^  are  in  the  same  relation  as  D 
and  D.  We  have  thus  proved,  in  the  case  in  which  D  and  D 
are  central  conies,  the  following  theorem. 

THEOREM  3.  If  the  directrices  of  two  cylinders  (with  parallel 
rulings')  are  similar  and  similarly  placed  conies,  or  if,  in  the  case 
of  hyperbolas,  each  is  similar  and  similarly  placed  either  to  the 
other  or  to  the  conjugate  of  the  other,  then  the  sections  of  the 
cylinders  by  the  same  plane  or  by  two  parallel  planes  stand  in  like 
relationship. 

The  converses  of  Theorems  1,  2,  3  are  true,  as  is  readily  seen. 
The  three  theorems  and  their  converses  can  be  stated  equally 
well  in  terms  of  projections.  Thus  Theorem  1  and  its  con- 
verse are  equivalent  to  the  theorem  :  A  plane  curve  is  a  conic 
of  a  certain  type  if  and  only  if  its  projection  on  a  plane  not  per- 
pendicular to  its  plane  is  a  conic  of  this  type. 


536  ANALYTIC   GEOMETRY 

EXERCISES 

1.  Do  Exs.  9,  10,  11  of  Ch.  XIX,  §  1. 

2.  Do  Exs.  20,  21,  22  of  Ch.  XIX,  §  1. 

What  does  each  of  the  following  equations  represent  ? 

3.  4z2  +  i/2  —  8x  +  4y  —  4  =  0. 

4.  3o2  +  6o;-2?/  +  l  =  0. 

5.  xy  +  2x-y  —  6  =  0. 

6.  Prove  Theorem  1  when  D  is  a  parabola. 

7.  Prove  Theorem  3  when  D  is  a  parabola. 

8.  State  Theorems  2  and  3  and  their  converses  in  terms  of 
projections. 

9.  Show  that  if  a  central  conic  S  in  a  plane  M  projects  into 
the  central  conic  D  in  the  plane  A",  then  the  center  0'  of  S 
projects  into  the  center  0  of  D. 

7.  Cones.  Let  a  plane  curve,  not  a  straight  line,  and  a 
point  0,  not  in  the  plane  of  the  curve,  be  given.  Draw  an 
indefinite  straight  line  through  0  and  each  point  of  the  curve. 
The  surface  formed  by  these  lines  is  known  as'  a  cone.  The 
lines  are  its  rulings,  or  generators,  and  the  point  0  is  its  vertex. 

If  the  given  curve  is  a  circle  and  0  lies  on  the  line  L 
through  its  center  perpendicular  to  its  plane,  the  cone  can  be 
generated  by  the  rotation  about  L  of  any  ruling.  Accordingly, 
it  is  known  as  a  cone  of  revolution  or  a  circular  cone.  The  line 
L  is  its  axis  and  the  constant  angle  between  L  and  a  ruling 
is  the  generating  angle. 

Problem.  To  find  the  equation  of  a  cone  of  revolution  whose 
vertex  is  at  the  origin,  whose  axis  is  the  axis  of  z,  and  whose 
generating  angle  is  </». 

Let  P :  (x,  y,  z)  be  any  point  of  the  cone  other  than  O.  The 
ruling  K  on  which  P  lies  determines  with  the  axis  of  z  a 
plane  M  which  cuts  the  (x,  ?/)-plane  in  a  line  L.  Direct  the 
line  L  as  shown  in  the  figure  and  denote  the  projection  of  OP 


SPHERES  AND  OTHER  SURFACES 


537 


on  L,  thus  directed,  by  r.  Then  the  directed  line  L  and  the 
axis  of  z  form  in  the  plane  M  a  system  of  coordinate  axes, 
with  respect  to  which  the  point  P  has 
the  coordinates  (r,  z). 

The  equation  of  the  ruling  R,  as  a  line 

of  M,  is ' 

r  =  z  tan  </>. 

Since  P  lies  on  R,  its  coordinates  (r,  z) 
satisfy  this  equation.     But,  clearly, 


r= 

where  the  plus  sign  or  the  minus  sign  is 

to  be  taken,  according  as  P  lies  on  the 

upper  nappe  of  the  cone  (as  shown)  or 

on  the  lower  nappe.     Consequently,  the  coordinates  (x,  y,  z) 

of  P  satisfy  the  equation  * 


FIG.  3 


!  =  z  tan  <£, 
or 

Conversely,  every  point  whose  coordinates  satisfy  this  equa- 
tion lies  on  the  cone,  for  the  steps  can  be  retraced.  Hence 
this  is  the  equation  of  the  cone. 

Equation  (1)  is  homogeneous  in  x,  y,  z.  This  is  characteristic 
of  the  equation  of  a  cone  with  its  vertex  at  the  origin.  In  fact, 
we  can  state  the  theorem :  A  curved  surface  is  a  cone  with  its 
vertex  at  the  origin,  when  and  only  when  its  equation  is  homogeneous 
in  x,  y,  z. 

Before  giving  the  proof  we  consider  a  particular  homo- 
geneous equation 

(2)  4  x2  —  3  y1  +  12  z2  =  0. 

If  ccj,  ylt  zl  is  a  solution  of  this  equation,  i.e.  if 

(3)  4aj1*-3y1'  +  12Zl*  =  0, 

*  The  coordinates  of  the  origin,  originally  ruled  out,  clearly  satisfy  the 
equation. 


538 


ANALYTIC   GEOMETRY 


kx1}  kyi,  kzlf  where  k  is  any  constant,  is  also  a  solution.     For, 
the  equation 


=  0    or 


-  3  ^  +  12  zfi  =  0 


is  true,  inasmuch  as  the  parenthesis  has,  by  (3),  the  value  zero. 
It  follows  that,  if  P  :  (x1}  yly  z^)  is  an  arbitrary  p<5int,  not 
the  origin,  on  the  surface  represented  by  (2),*  any  point  with 
coordinates  .of  the  form  (kxly  kyl}  kz^)  is  also  on  the  surface. 
But  the  points  (kxl9  kyl}  kz^,  where  k  is  an  arbitrary  constant, 
are  all  the  points  of  the  line  OP  passing  through  the  origin 
and  P  (Ex.  6  at  the  end  of  Ch.  XVIII).  That  is,  the  line  OP 
through  the  origin  lies  wholly  on  the  surface.  But  P  was  an 
arbitrary  point  on  the  surface,  other  than  0,  and  therefore  the 
surface  is  formed  by  lines  through  the  origin,  i.e.  it  is  a  cone 
with  its  vertex  at  the  origin. 

The  cone  can  be  constructed  by  drawing  the  curve  in  which 
it  intersects  a  plane  not  passing  through  0  and  by  joining  0 

with  the  points  of   this  curve   by 
straight    lines.      If    y  =  1    is    the 
plane  taken,  the  curve  of  intersec- 
~v   tion  is  the  ellipse 


FIG.  4 


and  thus  the  cone  is  as  shown  in 
Fig.  4. 

This  example  suggests  the  following  proof  for  the  theorem. 

An  equation  in  x,  y,  z  is  homogeneous  if  and  only  if,  when 
xi>  y\i  z\  (n°t  all  zero)  is  a  solution,  kx1}  kyl}  kzi}  where  k  is  an 
arbitrary  number,  is  also  a  solution.  On  the  other  hand,  a 
curved  surface  is  a  cone  with  its  vertex  at  the  origin  if  and 
only  if,  when  P :  (xi}  yly  Zj)  is  a  point  of  the  surface  other  than 
0,  an  arbitrary  point,  (kx1}  ky1}  kz^),  of  OP  is  also  a  point  of  the 
surface.  Thus  the  algebraic  condition  that  an  equation  be 
homogeneous  is  equivalent  to  the  geometrical  condition  that  a 


*  The  locus  of  (2)  is  actually  a  surface  (and  not  a  curve  or  a  point), 
since  all  pairs  of  values  (x,  z)  lead  to  points  on  it. 


SPHERES  AND   OTHER   SURFACES  539 

curved  surface  be  a  cone  with  its  vertex  at  the  origin,  and  the 
theorem  is  proved. 

Quadric  Cones.  A  cone  represented  by  a  homogeneous  equa-' 
tion  of  the  second  degree  in  #,  y,  z,  i.e.  by  an  equation  of  the 
form, 

(4)  ax2  +  by2  +  cz*  +  dxy  -f  eyz  +  fzx  =  0, 

is  called  a  quadric  cone. 

We  state,  without  proof,  that  equation  (4),  if  it  represents 
a  •  cone,*  can  always  be  transformed  by  a  rotation  of  axes 
(cf.  Ch.  XXIV,  §  6)  into  the  equation, 

(5)  Ax2  +  By2  -  Cz"-  =  0, 
where  A,  B,  and  C  are  positive  constants. 

The  quadric  cone  (5)  is,  in  particular,  a  cone  of  revolution, 
when  and  only  when  it  can  be  written  in  the  form  (1),  i.e. 
when  and  only  when  A  =  B. 

All  quadric  cones  are  of  one  general  type.  They  cannot  be 
classified  into  three  types,  corresponding  to  those  for  quadric 
cylinders.  For  ellipses,  hyperbolas,  and  parabolas  can  all  be 
obtained  as  plane  sections  of  any  one  of  them.f 

EXERCISES 

Construct  the  cones  represented  by  the  following  equations. 
If  the  cone  is  circular,  determine  its  axis  and  the  generating 
angle. 

1.  cc2+2/2-z2  =  0.  4.   4  a2  +  2/2  —  4z2  =  0. 

2.  <c2-3#2  +  z2  =  0.  5.   60?  —  3y*  —  2z*  =  0. 

3.  4z2-/2  —  z2=0.  6.     2 


*  Equation  (4)  represents  in  general  a  curved  surface,  and  hence,  by 
the  theorem,  a  cone.  Under  special  conditions  it  may,  however,  repre- 
sent two  planes,  a  single  plane,  a  line,  or  merely  the  origin.  For  example, 
x2  —  y*  =  0  represents  two  planes  ;  (x  —  y)2=  0,  a  single  plane  ;  x2  +  j/2  =  0, 
a  line  ;  and  x2  -f  y2  +  z2  =  0,  only  the  origin.  These  cases  are  here  excluded. 

t  This  was  proved  geometrically  for  the  case  of  a  cone  of  revolution  in 
Ch.  VIII,  §  10.  An  analytical  proof  covering  all  cases  will  be  given  later, 
Ch.  XXIII,  §  6. 


540  ANALYTIC   GEOMETRY 

Find  the  equations  of  the  following  cones. 

7.  The  cone  of  revolution  whose  vertex  is  at  the  origin, 
whose  axis   is   the   axis  of    y,  and   whose   generating  angle 
is  30°. 

8.  The  cone  of  revolution  whose  vertex  is  at  (0,  0,  a), 
whose  axis  is  the  axis  of  z,  and  whose  generating  angle  is  45°. 

Ans.     xz  +  y*—(z  —  a)2  =  0. 

9.  The  quadric  cone  which  has  its  vertex  at  the  origin  and 
intersects  the  plane  z  =  1  in  the  ellipse  whose  center  is  on  the 
z-axis,  whose  transverse  axis  is  parallel  to  the  axis  of  y,  and 
whose  major  and  minor  axes  are  6  and  4. 

10.  The  quadric  cone  which  has  its  vertex  at  (0,  1,  0)  and 
intersects  the  (z,  <c)-plane  in  the  hyperbola  2  a?  —  z2=  4. 

11.  The  cone  of  revolution  which  has  the  line  bisecting  the 
angle  between  the  positive  y-  and  z-axes  as  axis  and  which 
contains  the  y-  and  z-axes.  Ans.     x2  =  2  yz. 

12.  The  cone  of  revolution  which  has  the  line  x=y  =  z  as 
axis  and  passes  through  the  coordinate  axes. 

Ans.     xy  +  yz  +  zx  =  0. 

8.  Surfaces  of  Revolution.  Let  a  plane  curve  and  a  line  L  in 
the  plane  of  the  curve  be  given.  The  surface  generated  by  the 
curve  when  the  plane  is  rotated  about  L  through  360°  is  known 
as  a  surface  of  revolution.  The  line  L  is  its  axis. 

It  is  clear  that  spheres  are  surfaces  of  revolution.  So  also  are 
circular  cylinders  and  circular  cones  ;  in  the  one  case  the  gener- 
ating curve  is  a  line  parallel  to  L,  and  in  the  other,  it  is  a  line 
which  intersects  L. 

Quadric  Surfaces  of  Revolution.  The  surfaces  obtained  by 
rotating  the  conies  about  their  axes,  together  with  the  spheres 
and  the  circular  cylinders  and  cones,  are  known  as  quadric 
surfaces  of  revolution. 

Problem.  To  find  the  equation  of  the  ellipsoid  of  revolution 
generated  when  the  ellipse  in  the  (y,  z)-plane,  whose  equation 


SPHERES  AND  OTHER  SURFACES 


541 


in  that  plane  is 


is  rotated  about  its  conjugate  axis,  the  axis  of  z, 

Let  M  be  the  rotating  plane  in  an  arbitrary  position,  and  let 
P  :  (x,  y,  z)  be  an  arbitrary  point  on 
the  ellipse  in  M.  Establish  in  M  the 
same  system  of  axes  as  was  set  up  in 
the  plane  M  of  Fig.  3  in  finding  the 
equation  of  a  cone  of  revolution.  The 
equation  of  the  ellipse  in  M,  referred 
to  these  axes,  is 


ZL+1=1. 


FIG.  5 


Since  P  lies  on  the  ellipse,  its  coordinates  (r,  z)  satisfy  this 
equation.     Consequently,  inasmuch  as 

r2  =  x*  +  y\ 
the  coordinates  (x,  y,  z)  of  P  satisfy  the  equation 

(1)  —2  +  ^  +  ^  =  1,  a  >  6, 

and  this  is  the  equation  of  the  ellipsoid  of  revolution. 

In  a  similar  manner,  the  equation  of  the  ellipsoid  of  revolu- 
tion obtained  by  rotating  the  ellipse  about  its  transverse  axis, 
the  axis  of  y,  is  found  to  be 


(2) 


a>b. 


The  first  of  the  two  ellipsoids  of  revolution  (Fig.  6)  is  often 
called  an  oblate  spheroid  ;  and  the  second  (Fig.  7),  a  prolate 
spheroid.  Both  approach  as  ttteir  limits  the  sphere,  whose 
center  is  at  the  origin  and  whose  radius  is  a,  when  6  is  made 
to  approach  a  as  its  limit. 

The  hyperboloids  of  revolution  generated  when  the  hyperbolas 


(3) 


62 


542 


ANALYTIC   GEOMETRY 


situated  in  the  (y,  z)-plane,  are  rotated  about  the  axis  of  z,  —  the 
conjugate  axis  of  the  first  hyperbola,  and  the  transverse  axis 
of  the  second,  —  have  respectively  the  equations : 


(4) 
(5) 


|_  j . 

a?     a2     62 


?L  4.  Vl  _  ?!  =  _ 


6^ 


1, 


or 


_  ur-  _  */-      z-  __  . 
a2     a2     &2 


FIG.  6 


FIG.  7 


The  two  hyperboloids  of  revolution  are  shown  in  Figs.  8  and 
9.  The  first  is  known  as  a  hyperboloid  of  one  sheet  or  an 
imparted  hyperboloid ;  the  second,  as  a  hyperboloid  of  two  sheets, 
or  a  biparted  hyperboloid. 

Taken  together,  the  hyperboloids  (4)  and  (5)  are  known  as 
conjugate  hyperboloids  of  revolution.  They  are  generated  by 


FIG.  8 


FIG.  9 


the  conjugate  hyperbolas  (3)  revolving  about  the  same  axis,  the 
axis  of  z.     The  cone  which  results  from  the  rotation  about  this 


SPHERES  AND   OTHER  SURFACES 


543 


axis  of  the  common  asymptotes  of  the  hyperbolas,  namely,  the 
cone 


is  called  the  common  asymptotic  cone  of  the  conjugate  hyper- 
boloids.     See  Ch.  XXIII,  Fig.  5. 

Since  a  parabola  has  but  one  axis,  there  can  be  obtained 
from  it  but  one  quadric  surface,  or  paraboloid,  of  revolution, 
namely,  that  which  results  from  rotating 
it  about  its  axis.  If  the  equation  of  the 
parabola,  in  the  (y,  z)-plane,  is 

y2  =  2  raz, 

the  equation  of  the  paraboloid  of  revolution 

is 

(7)  x*  +  y2=-2mz. 

An  ellipsoid  or  hyperboloid  of  revolu- 
tion is  symmetric  in  the  center,  0,  of  the 
ellipse  or  hyperbola  which  generates  the 
surface.  Accordingly,  we  call  0  the  center 
of  the  surface. 

Every  surface  of  revolution  is  symmetric 
in  its  axis  and  in  every  plane  passing  through  the  axis.  An 
ellipsoid  or  hyperboloid  of  revolution  is  also  symmetric  in  the 
plane  through  the  center  perpendicular  to  the  axis,  and  in 
every  line  through  the  center  lying  in  this  plane.  Thus  the 
surface  (1)  is  symmetric,  not  only  in  the  axis  of  z  and  in  all 
planes  through  this  axis,  but  also  in  the  (x,  t/)-plane  and  all 
lines  lying  in  this  plane  and  passing  through  0. 

EXERCISES 

Establish  each  of  the  following  equations. 

1.    Equation  (2).          2.    Equation  (4).  3.    Equation  (5). 

What  surface  does  each  of  the  following  equations  represent  ? 
Construct  the  surface. 


Fio.  10 


544           »               ANALYTIC  GEOMETRY 

4.  x'i  +  y*  +  4z2  =  4.  8.   3a;2—  4y2  —  4z2=12. 

5.  4z2  +  9</2  +  9z2=36.  9.   3a?  -5y*  +  3z2  =  15. 

6.  2«2  +  3y2  +  2z2  =  6.  10. 

7.  a2  +  y2  —  2z2  =  2.  11. 


12.  The   parabola  of  the  text   is   rotated   about   the   axis 
of  y.     Find  the  equation  of  the  surface  generated  and  con- 
struct it. 

13.  The   surface   generated   by  the  rotation  of  a  circle  of 
radius  a  about  a  line  L  in  the  plane  of  the  circle  at  the  dis- 
tance b  >  a  from  its  center  is  called  an  anchor  ring,  or  torus. 
Find  its  equation,  if  L  is  the  axis  of  z  and  the  circle  is  in  the 
(y,  z)-plane  with  its  center  on  the  axis  of  y. 

Ans.    (  Vo;2  +  f  —  6)2  +  z2  =  a?. 

EXERCISES  ON   CHAPTER  XXII 

1.  Find  the  equation  of  the  sphere  having  the  line-segment 
joining  the  two  points  (3,  2,  —  1),  (5,  4,  3)  as  a  diameter. 

2.  Find  the  equation  of  the  sphere  which  has  its  center  at 
the  point  (5,  —2,  3)  and  is  tangent  to  the  plane3a;-|-2y-f  z=0. 

3.  A  sphere  has  its  center  in  the  plane  x  +  y  +  3z  —  2  =  0 
and  passes   through   the  three   points    (2,  3,  1),    (2,  —  1,  5), 
(—2,  —  3,  3).     Find  its  equation. 

4.  Find  the  equation  of  the  sphere  which  has  its  center  on 
the  line  4o;  +  8  =  3y  +  7  =  4z  and  passes  through  the  points 
(4,  3,  -  1),  (3,  2,  3). 

5.  A  sphere  is  tangent  to  the  plane  #  —  2y  —  2z  =  7  in  the 
point  (3,   —  1,   —  1)  and  goes  through  the  point  (1,  1,  —  3). 
Find  its  equation.          Ans.    xz  +  y2  +  z1  —  10  y  —  10  z  —  31  =  0. 

6.  There   are   two   spheres    passing    through    the    points 
(4,  0,  3),  (5,  4,  0),  (5,  1,  3)  and  having  the  radius  3.     Find  their 
equations. 

f    x*+    y*+    z2—    6x—  4y  —  2z+    5  =  0. 


SPHERES  AND   OTHER   SURFACES  545 

7.  Find  the  equations  of  the  spheres  which  are  tangent  to 
the  plane  x  +  2y  —  2z  —  12  =  0  and  pass  through  the  three 
points  (3,  -  2,  0),  (2,  -  3,  0),  (3,  1,  -  3). 

8.  Find  the  equations  of  the  spheres  which  are  tangent 
to  the  planes  2x —  y +  2z  +  1  =  0,  Qx  +  3y  —  2z  +  5  =  0  and 
have  their  centers  on  the  line  1  —  a?  =y  -f  1  =  2  z. 

9.  Find  the  equation  of  the  sphere  inscribed  in  the  tetra- 
hedron  formed   by   the   plane   2x  +  2y  +  z  —  4  =  0   and    the 
coordinate  planes. 

10.  A  sphere  goes  through  the  point  (4,  6,  3)  and  meets  the 
(x,  y)-plane  in  a  circle  whose  center  is  at  the  point  (1,  2,  0) 
and  whose  radius  is  5.     Find  its  equation. 

Ans.   z2  +  ?/2 +  z2-2z-4?/-3z- 20  =  0. 

11.  Find  the  equations  of  the  tangent  line  to  the  circle 
a;2  +  y*  +  z~  —  z  +  4z  =  0,     3z-2?/  +  4z  +  l  =  0 

at  the  point  (1,  -  2,  -  2). 

12.  Does  the  line  2x  —  l=y+3=4—  z  intersect  the  sphere 

«2  +  2/2  -I-  z-  —  Gx  +  8 y  —  4z  +  4  =  0 ? 

13.  Show  that  the  radical  planes  of  three  spheres,  taken  in 
pairs,*  pass  through  a  line  or  are  parallel. 

ORTHOGONALITY  f 

14.  Prove  that  the  plane  x  +  y  +  z  —  1  =  0  intersects  the 
sphere  of  Ex.  12  orthogonally. 

15.  When  does  the  plane  ax  +  by  +  cz  +  d  =  Q  intersect  the 
sphere 

a2  +  yz  +  z2  +  Ax  +  By  +  Cz  +  D  =  0 

orthogonally  ? 

Ans.   When  and  only  when  a  A  +  bB  +  cC  =  2  d. 

*  It  is  assumed  that  no  two  of  the  three  spheres  are  concentric ;  cf . 
footnote,  p.  530. 

t  Cf.  Exs.  18-26  at  the  end  of  Ch.  IV. 


546  ANALYTIC   GEOMETRY 

16.  Show  that  the  line  3z  +  8  =  —  6y  —  7  =  2z  +  13  inter- 
sects the  sphere  of  Ex.  12  orthogonally. 

17.  Find  the  condition  that  the  line 


dx  =  0,        020;  +  b?y  +  c2z  +  d2  =  0 
intersect  the  sphere  of  Ex.  15  orthogonally. 

18.  Find  the  angle  of  intersection  of  the  spheres  (2)  of  §  5. 

19.  Prove  that  the  two  spheres 

a2  +  y2  +  z2  -  9  =  0,         z2  +  /  +  z2-6a;  +  82/  +  9  =  0 
intersect  orthogonally. 

20.  When  does  the  sphere  of  Ex.  15  intersect  the  sphere 


orthogonally  ? 

Ana.   When  and  only  when     AA'  +  BB'  +  CC'  =  2D+2D'. 

21.  Find  the  equation  of  the  sphere  containing  the  circle  of 
Ex.  11  and  intersecting  orthogonally  the  first  of  the  spheres 
of  Ex.  19. 

Loci 

22.  Do  Ex.  1  of  Ch.  V,  §  1,  when  P  is  not  restricted  to  lie 
in  a  plane. 

23.  The  same  for  Ex.  4  of  Ch.  V,  §  1. 

24.  A  point  P  moves  so  that  the  ratio  of  its  distances  from 
two  fixed  points  is  constant.     Find  its  locus. 

25.  A  point  P  moves  so  that  its  distance  from  a  given  plane 
M  is  proportional  to  the  square  of  its  distance  to  a  given  point 
P0,  not  in  M.     If  P  remains  always  on  the  same  side  of  M  as 
P0,  find  its  locus. 

26.  If,  in  the  preceding  exercise,  P0  lies  in  M  and  P  may  be 
on  either  side  of  M,  what  is  the  locus  of  P? 

27.  What  is  the  locus  of  a  point  which  moves  so  that  its 
distance  from  a  given  line  is  proportional  to  its  distance  from 
a  given  plane  perpendicular  to  the  line  ? 


SPHERES  AND    OTHER  SURFACES  547 

28.  What  is  the  locus  of  a  point  which  moves  so  that  its 
distance  from  a  given  point  is  proportional  to  its  distance  to  a 
given  plane  through  the  point  ? 

29.  What  is  the  locus  of  a  point  P  which  moves  so  that  the 
difference  of  the  squares  of  its  distances  from  a  given  point 
and  a  given  sphere  is  constant,  if  the  distance  from  P  to  the 
sphere   is   measured   along    a    tangent    line    to    the    sphere 
through  P? 

30.  A  given  point  P0  is  distant  2  a  units  from  a  given  plane 
and  P'  is  an  arbitrary  point  in  the  plane.     What  is  the  locus 
of  the  point  P  so  chosen  on  the  line  P0P'  that  P0P  •  P0P'  =  4 a2? 

Suggestion.      Take    the    coordinates    of    P'    as    auxiliary 
variables. 

31.  A  is  a  fixed  point  and  R  an  arbitrary  point  of  a  given 
sphere  whose  center  is  0.     The  radius  OA  is  produced  four 
times  its  length  to  the  point  A'  and  the  radius  OR,  twice  its 
length  to  the  point  R'.     What  is  the  locus  of   the  point  of 
intersection  of  AR  and  A'R'  ? 


CHAPTER  XXIII 


QUADRIC   SURFACES 

1.  The  Ellipsoid.  A  quadric  surface  is  any  surface  defined 
by  an  equation  of  the  second  degree  in  x,  y,  z.  The  sphere  and 
the  quadric  cylinders,  cones  and  surfaces  of  revolution  studied 
in  the  previous  chapter  are  special  types  of  quadric  surfaces. 
We  proceed  to  consider  more  general  types. 

The  surface  defined  by  the  equation 

(1)  /'  l  +  l  +  5  =  1      -;;;' 

is  known  as  an  ellipsoid.  If  two  of  the  three  numbers  a,  &,  c 
are  equal,  it  is  in  particular  an  ellip- 
soid of  revolution  (Ch.  XXII,  §  8). 
To  construct  the  surface  in  the  general 
case  when  no  two  of  the  three  numbers 
a,  &,  c  are  equal,  plot  first  the  sections 
by  the  coordinate  planes,  that  is,  the 
curves  of  intersection  with  the  planes 
x  =  0,  y  =  0,  z  =  0.  These  are,  respec- 
tively, the  ellipses 

x2     z2 ..          x1  .  y*  _  i 


Fia.  1 


&2      c2 


The  parts  of  these  ellipses  which  lie  in  the  quarter-planes 
bounding  the  first  octant  connect  the  points  (a,  0,  0),  (0,  6,  0), 
(0,  0,  c),  as  shown. 

The  section  of  (1)  by  a  plane  z  =  k  parallel  to  the  (x,  y)- 
plane  has  the  equations  : 

548 


QUADRIC   SURFACES 


549 


£?_l_£=l_fc? 


=  k. 


If  kz  <  c2,  these  equations  represent,  in  the  plane  z  =  k,  an 
ellipse  whose  center  is  on  the  axis  of  z  and  whose  axes  lie  in 
the  (z,  x)-  and  (y,  z)-planes  and  have  the  lengths 


2a\  1-- 


c2 


FIG.  2 


As  A;  increases  from  0  toward  c  as  its  limit,  this  ellipse,  rising 
from  the  section  by  the  (x,  ?/)-plane,  grows  continuously 
smaller  and  shrinks  finally  to  a  point,  —  the  point  (0,  0,  c). 
Similarly,  if  k  decreases  from  0  toward  —  c  as  its  limit. 

The  surface  generated  by  the 
changing  ellipse  is  the  ellipsoid. 
Fig.  2  (or  Fig.  6  of  Ch.  XXII  *) 
shows  it  in  its  entirety.  The  sur- 
face is  evidently  symmetric  in  the 
origin,  0,  and  in  the  coordinate 
axes  and  coordinate  planes.  0  is 
called  the  center  of  the  ellipsoid ; 
the  coordinate  axes,  the  axes  of  the 
ellipsoid  ;  and  the  coordinate  planes,  the  principal  planes  of  the 
ellipsoid.  The  sections  by  the  principal  planes  are  known  as 
the  principal  sections. 

The  dimensions  of  the  ellipsoid,  measured  along  the  axes, 
are  2a,  26, '2c.  These  numbers,  in  the  order  of  their  magni- 
tude, are  known  as  the  major  axis,  mean  axis,  and  minor  axis 
of  the  ellipsoid. 

Here,  and  throughout  the  chapter,  we  speak  of  a  (plane) 
section  of  a  surface  only  when  the  plane  in  question  meets  the 
surface  in  a  curved  line.  Sections  by  parallel  planes  we  shall 
call  parallel  sections. 

*  Figs.  6-10  of  Ch.  XXII,  drawn  originally  to  represent  quadric  sur- 
faces of  revolution,  picture  equally  well  the  corresponding  general  quad- 
ric surfaces  studied  in  this  chapter.  One  has  merely  to  imagine  that  a 
different  ratio  of  foreshortening  along  the  axis  of  x  has  been  chosen. 


550 


ANALYTIC   GEOMETRY 


EXERCISES 


Construct  the  following  ellipsoids,  drawing  accurately  the 
principal  sections  and  the  sections  parallel  to  one  principal 
plane.  What  are  the  lengths  of  the  axes  ? 


352 


**  _  1 


2.   9  z2  +  36  y2  +  4  z2  =  36. 


3.  Discuss  the  generation  of   the  ellipsoid  (1)  by  sections 
parallel  to  the  (z,  x)-plane. 

4.  Prove  that  the  sections  of  the  ellipsoid  (1),  which  are 
parallel  to  a  principal  plane,  e.g.  the  (x,  y)-plane,  are  similar 
and  similarly  placed  ellipses  ;  cf.  p.  260. 


2.   The  Hyperboloids.     The  Hyperboloid  of  One  Sheet. 
surface  represented  by  the  equation 


The 


(1) 


&2 


is  called  a  hyperboloid  of  one  sheet  or  an  imparted  hyperboloid. 

If  a  =  6,  it  is  in  particular  a  hyper- 
boloid of  revolution  of  one  sheet 
(Ch.  XXII,  §  8). 

In  the  general  case,  a  ^=  b,  the 
surface  can  be  constructed  by  the 
method  of  §  1.  The  sections  by 
the  vertical  coordinate  planes, 
x  =  0  and  y  =  0,  are  the  hyper- 
bolas 


FIG.  3 


62 


—  —  -=1 

a2      c2" 
The  smallest 


The  sections  by  the  planes  z  =  k  are  ellipses. 
one  is  the  section  by  z  =  0,  the  (x,  ?/)-plane ;  it  is  known  as 
the  minimum  ellipse.  The  general  one  increases  in  size  as  its 
distance  from  the  (x,  y)-plane  increases.  The  surface  can  be 
thought  of  as  generated  by  it ;  cf.  Fig.  8,  Ch.  XXII. 


QUADRIC   SURFACES 


551 


The  Hyperbola  id  of  Two  Sheets.     This  surface,  also  known 
as  the  biparted  hyperboloid,  is  denned  by  the  equation 


(3) 


—  -I-  ^ =  —1     or     — 


a2 


A  particular  case,  when  a  =  b,  is  the  hyperboloid  of  revolution 
of  two  sheets. 

In  the  general  case,  a  =£  6,  the  sections  by  the  vertical  coor- 
dinate planes  are  the  hyperbolas  conjugate  to  the  hyperbolas 
(2).  The  (x,  y)-plane,  z  =  0,  does  not 
intersect  the  surface.  This  is  true 
of  all  the  planes  z  =  k,  for  which 
k2  <  c2.  The  planes  z  =  ±  c  meet  the 
surface  in  the  points  (0,  0,  ±  c),  and 
the  planes  z  =  k,  where  kz  >  c2,  meet 
it  in  ellipses,  which  increase  in  size 
as  k  increases  in  numerical  value  ; 
cf.  Fig.  9,  Ch.  XXII. 


FIG.  4 


Center,  Axes,  Principal  Planes. 
Each  hyperboloid  is  symmetric  in 
the  origin  0  and  in  the  coordinate 
axes  and  coordinate  planes ;  0  is  the  center,  the  coordinate 
axes,  the  axes,  and  the  coordinate  planes,  the  principal  planes 
for  each  surface.  The  sections  by  the  principal  planes  are  the 
principal  sections. 

The  Asymptotic  Cone.  The  hyperboloids  (1)  and  (3)  are 
called  conjugate  hyperboloids.  We  have  seen  that  each  vertical 
coordinate  plane  intersects  them  in  conjugate  hyperbolas  whose 
common  asymptotes  pass  through  the  origin.  This  is  true  also 
of  any  vertical  plane, 
(4)  y  =  mx, 

which  passes  through  the  axis  of  z.  For,  the  sections  of  (1) 
and  (3)  by  the  plane  (4)  are  also  the  sections  by  this  plane  of 
the  cylinders, 

O^!  +  ™M_-  = 
c2 


552 


ANALYTIC   GEOMETRY 


whose  equations  are  obtained  by  eliminating  y  from  (4)  and 
(1),  and  from  (4)  and  (3).  But  these  cylinders  are  conjugate 
hyperbolic  cylinders  whose  common  asymptotic  planes  are 


(5) 


Consequently,  by  Ch.  XXII,  §  6,  Th.  2,  their  sections  by  the 
plane  (4)  are  conjugate  hyperbolas  whose  common  asymptotes, 
denned  by  equations  (4)  and  (5),  pass  through 
the  origin,  q.  e.  d. 

The  equation  of  the  locus  of  these  asym- 
ptotes, as  the  plane  (4)  rotates  about  the  axis 
of  z,  is  obtained  by  eliminating  ra  (now  an 
auxiliary  variable  expressing  the  motion  of 
the  plane)  from  equations  (4)  and  (5).  The 
result  is 


(6) 


a2     b2     c2 


FIG.  5 


The  locus  of  the  asymptotes  is,  therefore,  a 
cone  whose  vertex  is  at  the  origin.  This 
cone  is  called  the  asymptotic  cone  of  each  of  the  hyperboloids 
(1)  and  (3).  Evidently  (1)  lies  wholly  without  the  cone,  that 
is,  on  the  convex  side  of  it,  while  (3)  lies  wholly  within  it. 


EXERCISES 


Construct  the  following  hyperboloids,  drawing  accurately 
the  principal  sections  which  exist -and  the  sections  parallel  to 
one  principal  plane. 


i     * i  y_ "   — 

94      25 


3. 


2. 


— 
25 


-  9z2  =  36. 


5.  What  is  the  equation  of  the  hyperboloid  conjugate  to  the 
hyperboloid  of  Ex.  3  ?  of  Ex.  4  ?  Give  also  in  each  case  the 
equation  of  the  common  asymptotic  cone. 


QUADRIC   SURFACES 


553 


6.  Show  that  the  sections  of  the  hyperboloid  of  two  sheets 
(3),  which  are  parallel  to  a  vertical  principal  plane,  are  similar 
and  similarly  placed  hyperbolas. 

7.  The  sections  of  the  hyperboloid  of  one  sheet  (1)  by  two 
planes  parallel  to  a  vertical  principal  plane  are  similar  and 
similarly  placed  hyperbolas,  or  are  hyperbolas  each  of  which 
is  similar  and  similarly  placed  to  the  conjugate  of  the  other. 
Prove  this  theorem  and  determine  when  each  of  the  two  cases 
occurs. 

3.   The  Paraboloids.     Tlte  Elliptic  Paraboloid.     The  surface 
defined  by  the  e'quation 

1-2         ?/2 

/-\  \  ""    §  v       o  ~ 

(1)  h  —  =  -  2 

a2     b2 

is  called  an  elliptic  paraboloid.     If  a  =  b,  it  is  in  particular  a 
paraboloid  of  revolution  (Ch.  XXII,  §  8). 

In  the  general    case,  a  3=  b,  the =i 

sections  of  the  surface  by  the  verti 
cal  coordinate  planes,  y  =  0  and 
x  =  0,  are  the  parabolas 

both  of  which  open  upwards.     The 

(x,  y)-plane   intersects  the  surface 

only  in  the  origin.     A  plane  parallel 

to  the  (x,  ?/)-plane  and  below  it  does 

not  meet  the  surface,  while  a  plane 

parallel    to    the    (x,   y)-plane    and 

above   it  intersects  the  surface   in 

an  ellipse,  which  increases  in  size  as  the  height  of  the  plane 

increases ;  cf.  Fig.  10,  Ch.  XXII. 


Fia.  6 


The  Hyperbolic  Paraboloid. 
the  equation 

(2)  £- 

v  ;  * 


This  is  the  surface  defined  by 


554 


ANALYTIC   GEOMETRY 


It  is  never  a  surface  of  revolution,  no  matter  what  values  are 
assigned  to  a  and  6. 

The  sections  of  the  surface  by  the  vertical  coordinate  planes 
are  the  parabolas 


FIG.  7 


of  which  the  first  opens  upwards  and  the  second,  downwards. 

The  section  by  the  (x,  y)-plane 
consists  of  the  two  lines, 

OA:     *+2  =  0, 
a     b 

O  7?  •     x  —  ^  —  0 
un  .  —  —  u. 

a     b 

A  section  parallel  to  and  above 
the  (x,  y)-plane  is  a  hyperbola 
whose  vertices  are  on  the  parabola 
opening  upwards,  whereas  a  sec- 
tion parallel  to  and  below  the  (x,  y)-plane  is  a  hyperbola 
whose  vertices  are  on  the  parabola  opening  downwards.  It  is 
seen,  then,  that  the  surface  is  saddle-shaped;  it  rises  along  the 
parabola  which  opens  upwards,  and  falls  along  the  parabola 
which  opens  downwards.  The  (z,  o;)-plane  contains  the  pommel 
and  the  (y,  z)-plane,  the 
stirrups. 

The  surface  can  best 
be  plotted  by  drawing 
the  sections  parallel  to  a 
vertical  coordinate  plane, 
for  example,  the  (y,  z)- 
plane.  These  sections 
are  all  parabolas  opening 
downwards  and  having 
their  vertices  on  the  parabola  in  the  (z,  a)-plane.  Figure  8 
shows  part  of  the  surface  constructed  by  means  of  them. 

Vertex,  Axis,  Principal  Planes.     Each    paraboloid    is    sym- 
metric in  only  one  line,  the  axis  of  z,  and  in  only  two  planes, 


FIG.  8 


QUADRIC   SURFACES  555 

the  vertical  coordinate  planes.  The  line  is  known  as  the  axis, 
and  the  planes  as  the  principal  planes.  The  sections  by  the 
principal  planes  are  called  the  principal  sections,  and  the  point 
0,  the  vertex. 

EXERCISES 

Construct  accurately  the  following  paraboloids. 

1.    x-  +  y-  =  2z.  2.    ?-£  =  2z. 

94  94 

3.    2z2  +  322  =  12.  4.   z2-4i/2  =  —  82. 


5.  Prove  that  the  sections  of  a  paraboloid  of  either  type, 
which  are  parallel  to  a  principal  plane,  are  equal  and  similarly 
placed  parabolas. 

6.  Prove  that  the  elliptic  paraboloid  (1)  can  be  generated 
by  the  parabola  x2  =  2  a-z  moving  so  that  its  vertex  traces  the 
parabola  yz  =  2  b"z  in  x  =  0,  while  its  axis  remains  vertical  and 
its  plane  parallel  to  the  (z,  #)-plane. 

7.  Describe  and  prove  a  method  of  generating  the  hyper- 
bolic paraboloid,  which  is  similar  to  that  given  in  Ex.  6  for 
the  elliptic  paraboloid. 

8.  Show  that  the  equation  xy  =  az  represents  a  hyperbolic 
paraboloid. 

4.   Rulings.     The  Hyperboloid  of  One  Sheet.     The  equation 

CD  -i 

a- 
can  be  written  in  the  form 


Consider  the  equations 


H      —  -    =     --i 
b     c  a 


obtained  from  (2)  by  setting  the  first  factor  on  the  left  equal 
to  the  parameter  u  times  the  first  factor  on  the  right,  and 


556  ANALYTIC   GEOMETRY 

then  the  second  factor  on  the  right  equal  to  u  times  the  sec- 
ond factor  on  the  left. 

These  equations  represent  a  one-parameter  family  of  lines, 
each  line  being  given  by  a  particular  value  of  the  parameter  w. 
All  the  lines  lie  on  the  surface  (1).  For,  if  P  is  an  arbitrary 
point  of  the  line  u  =  u0)  the  coordinates  (x,  y,  z)  of  P  satisfy 
equations  U\  for  u  =  u0 ;  hence  they  also  satisfy  equation  (2), 
since,  if  u  =  MQ  is  eliminated  from  equations  Ui  by  multiplying 
them  together,  side  for  side,  the  result  is  precisely  equation  (2). 

There  will  be  just  one  line  t/i  through  an  arbitrary  point 
(o;0,  y0,  z0)  of  the  surface  (1)  if  the  equations 

(3)  &  +  5>= 

be 

have  a  unique  simultaneous  solution  for  u.  Let  us  see  when 
this  is  the  case. 

If  1  -f  x0/a  =f=  0,  the  first  equation  determines  u  uniquely, 
and  the  value  obtained  is  seen  to  satisfy  the  second,  since, 
when  it  is  substituted  in  the  second,  this  equation  takes  on 
the  form  (2)  for  (x,  y,  z)  ==  (x$,  y0,  ZQ}.  In  this  case,  then, 
there  is  just  one  line  C/i  through  (a:0,  y0,  z^). 

If  1  +  x0/a  =  0,  but  y0/b  —  z0/c  =£  0,  it  follows  from  (2)  that 
y0/b  +  z0/c  =  0.  Then  the  first  equation  of  (3)  is  satisfied,  no 
matter  what  value  u  has.  The  second  equation  determines 
u  uniquely,  and  so  in  this  case,  too,  there  is  just  one  line  C/i 
through  (XQ,  y0,  z0). 

Finally,  if  1  +  xQ/a  =  0  and  y0/b  —  z0/c  =  0,  at  least  one  of 
the  equations  (3)  is  contradictory  and  there  is  no  line  Ui 
through  (x0,  y0,  z0).  It  is,  however,  natural  to  supplement  the 
lines  C/i  by  the  line 

U0:  I--=°>  1  +  -  =  0, 

be  a 

for,  if  we  divide  each  of  the  equations  C/i  by  u  and  then  allow 
u  to  become  infinite,  the  line  t/i  approaches  U0  as  its  limit. 

We  have  proved,  then,  that  the  lines  U  consisting  of  the 
family  of  lines  t/i  and  the  line  U0  fill  out  the  surface  just 


QUADRIC   SURFACES 


557 


once.     They  form  what  is  called  a  set  of  rectilinear  generators 
or  rulings  of  the  surface. 

There  is  a  second  set  of  rulings,  V,  consisting  of  the  family 
of  lines 


b 

and  the  line 


1  +  -  =  0. 
a 


It  is  readily  seen  that  this  set  has  the  same  properties  as  the 
set  U.     Hence  we  have  the  theorem  : 

THEOREM  1.  A  hyperboloid  of  one  sheet  contains  two  sets  of 
rulings.  Through  each  point  of  the  surface  passes  one  ruling  of 
each  set. 

It  is  conceivable  that  the  ruling  U  and  the  ruling  V  which 
go  through  the  same  point  coincide.  This  is  not  the  case, 
however,  as  will  appear  later ;  *  cf.  Theorem  5. 

The  lines  through  the  origin  parallel  to  the  lines  U  have  the 
equations 

i»/  fy  /)* 

- +- =  U  — 

c         a' 


(4) 


rz\         x 
£__=__; 

cl         a 


S'-H 


Their  locus,  obtained  by  eliminating  u,  is  the  asymptotic  cone 

/v.2        iff        ?1 

- + y~  _  *. = o. 

a2      62      c2 

Moreover,  the  lines  fill  out  the  cone  just  once,  as  can  be  shown 
by  the  method  used  in  proving  Theorem  1. 

Similarly,  the  lines  through  the  origin  parallel  to  the. lines 
V,  i.e.  the  lines 

y  _z  __    x 
be        a' 

/y          g\  /£ 

[    \b     cj          a' 
*  It  is  not  difficult  to  give  a  direct  proof  of  the  fact  at  this  point. 


(5) 


b      c 


558  ANALYTIC   GEOMETRY 

also  fill  out  the  asymptotic  cone  just  once.     Consequently,  we 
have  proved  the  theorem  : 

THEOREM  2.  The  lines  ivhich  pass  through  the  center  of  a 
hyperboloid  of  one  sheet  and  are  parallel  to  the  rulings  U  (or  V} 
are  precisely  the  elements  of  the  asymptotic  cone.  In  other  words, 
there  is  one  and  only  one  ruling  of  each  set  which  is  parallel  to  a 
given  element  of  the  cone,  and  conversely. 

From  this  theorem  we  can  draw  the  following  conclusions. 

THEOREM  3.     No  three  rulings  of  one  set  are  parallel  to  a  plane. 

For  otherwise  there  would  be  three  elements  of  the  cone 
lying  in  a  plane,  and  this  is  impossible. 

THEOREM  4.  Two  rulings  of  one  set  neither  intersect  nor  are 
parallel;  that  is,  they  are  never  coplanar. 

For,  they  are  not  parallel,  since  no  two  elements  of  the  cone 
are  parallel  ;  and  they  do  not  intersect,  since  otherwise  there 
would  be  a  point  on  the  surface,  through  which  pass  two  rul- 
ings of  the  same  set. 

THEOREM  5.  Two  rulings  of  different  sets  either  intersect  or 
are  parallel;  that  is,  they  are  always  coplanar. 

For,  first,  the  rulings  of  the  two  sets  are  parallel  in  pairs, 
since  there  is  just  one  ruling  of  each  set  which  is  parallel  to 
a  given  element  of  the  cone.  From  equations  (4)  and  (5)  it 
appears  that  u  =  MO(=£  0)  and  v  =  v0(3=  0)  determine  a  pair  of 
parallel  rulings  if  and  only  if  1  +  U^VQ  =  0,  and  that  the  ruling 
u  =  0  is  parallel  to  F"0,  and  the  ruling  v  =  0  to  U0. 

Secondly,  two  non-parallel  rulings  of  different  sets  intersect 
in  just  one  point.  For,  it  is  easily  shown  that  the  four  equa- 
tions t/i  and  Fi*  which  define  in  pairs  two  rulings  which  are 
not  parallel,  i.e.  for  which  1  -f  uv  =£  0,  have  one  and  just  one 
simultaneous  solution  for  x,  y,  z,  namely, 


/c\ 
(6) 


1  —  uv  ,  u  4-  v  u  —  v 

—    -,     y  =  b—   -,     z  =  c  —   -, 

1  -J-  UV  1  +  UV  14-  UV 


*  The  proof  in  the  special  cases,  in  which  U0  or  F0  or  both  are  in- 
volved, is  left  to  the  student. 


QUADRIC   SURFACES  559 

If  u  and  v  take  on  all  possible  pairs  of  values  for  which 
1  +  uv  =£  0,  equations  (6)  give  the  coordinates  of  the  points  of 
intersection  of  all  the  lines  C/i  with  the  lines  V},  that  is,  the 
coordinates  of  all  the  points  of  the  surface  (1)  except  those 
on  the  lines  U0  and  V0.  They  constitute,  then,  a  parametric 
representation  of  the  surface  (1)  in  terms  of  the  two  parameters 
u  and  v. 

The  Hyperbolic  Paraboloid.     The  equation 

a2     52 
can  be  written  as 

(8) 


6, 

Accordingly,  there  are  tivo  sets  of  ridings  on  the  hyperbolic 
paraboloid,  namely: 

u'-  I+?-«' 

V-  ?_^  — 

a     b 

The  first  equation  of   U  represents  a  plane  parallel  to  or 
coincident  with  the  plane  AOz  (Fig.  8)  : 

a     b 

Consequently,  the  rulings  U  lie  one  each  in  the  planes  parallel 
to  (and  including)  the  plane  AOz.  Moreover,  they  are  the 
total  intersection  of  these  planes  with  the  surface ;  for,  if  the 
first  equation  of  U  is  solved  with  equation  (8)  of  the  surface, 
the  result  is  precisely  the  second  equation  of  U.  Similarly, 
the  rulings  Flie  one  each  in  the  planes  parallel  to  (and  in- 
cluding) the  plane  BOz : 


and  are  the  total  intersection  of  these  planes  with  the  surface. 
The  planes  AOz  and  BOz  are  known  as  the  directrix  planes. 


560  ANALYTIC   GEOMETRY 

There  is  just  one  ruling  of  each  set  through  each  point  of  the 
surface,  for  the  planes  parallel  to  (and  including)  a  directrix 
plane  exhaust  all  points  of  space  just  once,  and  hence  their 
lines  of  intersection  with  the  surface  exhaust  all  points  of  the 
surface  just  once. 

It  is  easily  shown  that  the  direction  components  of  a 
ruling  U  are  a,  —  6,  u,  and  that  those  of  a  ruling  F  are  a, 
6,  v.  Since  the  two  triples  are  never  proportional,  two  rulings 
of  different  sets  are  never  parallel  or  coincident.  In  particu- 
lar, the  two  rulings  which  pass  through  one  and  the  same  point 
of  the  surface  are  distinct. 

The  following  theorems  are  now  easily  proved. 

THEOREM  6.  Three  ridings  of  one  set  are  always  parallel  to 
a  plane. 

For,  all  the  rulings  of  a  set  are  parallel  to  a  directrix  plane. 

THEOREM  7.     Two  rulings  of  one  set  are  never  coplanar. 

For,  they  do  not  meet  since  they  lie  in  parallel  planes,  and 
they  are  not  parallel,  as  inspection  of  the  direction  components 
just  found  shows. 

THEOREM  8.     Two  rulings  of  different  sets  ahvays  intersect. 

For,  their  projections  on  the  (x,  y)-plane,  being  lines  in  the 
direction  of  OA  and  OB  respectively,  intersect  in  a  point  M. 
Now  there  is  but  one  point,  P,  on  the  surface  which  projects 
into  M,  since  a  line  perpendicular  to  the  (x,  i/)-plane  meets  the 
surface  just  once.  Consequently,  the  two  rulings  in  question 
intersect  at  this  point  P. 

The  coordinates  of  P,  found  by  solving  the  four  equations 
U  and  V  simultaneously  for  x,  y,  z,  are 

/m  U  +  V  ,  U  —  V  UV 

(9)  x  =  a— — ,         y  =  b  —    —          z  =  — 

2  22 

These  equations  constitute  a  parametric  representation  of  the 
surface  (7)  in  terms  of  the  parameters  u  and  v ;  there  are  no 
exceptional  points. 


QUADRIC   SURFACES  561 

EXERCISES 

1.  Find  the  equations  of  the  rulings  which  pass  through  the 
point  (3,  2,  5)  of  the  hyperboloid  of  one  sheet  of  Ex.  1,  §  2. 

2.  The  same  for  the  hyperbolic  paraboloid  of  Ex.  2,  §  3,  the 
point  on  the  surface  being  (9,  2,  4). 

Exercises  3-5.    Use  considerations  of  symmetry  in  the  proofs. 

3.  The  two  rulings  through  a  point  P  on  a  principal  section 
of  a  hyperboloid  of  one  sheet  are  equally  inclined  to  the  plane 
of  the  section  and  lie  in  a  plane  M  which  is  perpendicular  to 
the  plane  of  the  section. 

4.  The  same  for  a  hyperbolic  paraboloid. 

5.  If  P  and  P1  are  points  of  a  hyperboloid  of  one  sheet 
which  are  symmetric  in  the  center,  the  rulings  through  P  are 
parallel  to  those  through  P. 

6.  Assuming  Th.  1,  §  7,  prove  that  the  plane  M  of  Ex.  3 
passes  through  the  tangent  line  at  P  to  the  principal  section. 
Hence  show  that  the  projections  of  the  rulings  of  either  set 
on  a  principal  plane  are  the  tangents  to  the  principal  section 
in  that  plane. 

7.  The  same  for  a  hyperbolic  paraboloid,  applying  the  re- 
sults of  Ex.  4. 

8.  Prove  that  the  plane  determined  by  two  parallel  rulings 
of  a  hyperboloid  of  one  sheet  is  tangent  to  the  asymptotic  cone 
along  the  element  which  is  parallel  to  the  two  rulings. 

9.  Prove  that  there  are  no  straight  lines  on  (a)  an  ellipsoid ; 
(6)  a  hyperboloid  of  two  sheets  ;  (c)  an  elliptic  paraboloid. 

5.  Parallel  Sections.  Equations  (1),  (3),  and  (6),  §  (2),  of  a 
hyperboloid  of  one  sheet,  HI,  of  the  conjugate  hyperboloid  of 
two  sheets,.  H^  and  of  the  common  asymptotic  cone,  C,  can  be 
written  as  the  one  equation 


562  ANALYTIC   GEOMETRY 

where  A  is  given  the  values  1,  —  1,  and  0  in  turn.  The  three 
surfaces  can  thus  be  considered  simultaneously. 

We  propose  to  determine  the  sections  of  the  surfaces  (1)  by 
an  arbitrary  plane, 

(2)  Ax  +  By  +  Cz  =  k. 

At  least  one  of  the  coefficients  A,  B,  C  is  not  zero.  Assume 
that  (7  =£  0.  Then  the  sections  in  question  are  also  the  curves 
in  which  the  plane  (2)  meets  the  cylinders  whose  equations 
result  from  the  elimination  of  z  from  equations  (1)  and  (2), 
namely,  the  cylinders 


(3)  _---  = 

a2     62  c2<72 

Considering  the  sections  from  this  point  of  view,  we  conclude 
the  following  theorems. 

THEOREM  1.     The  section  of  a  hyperboloid  or  a  cone  is  a  conic. 
For,  the  cylinders  (3)  are  quadric  cylinders,  and  a  section 
of  a  quadric  cylinder  is  a  conic  ;  cf  .  Ch.  XXII,  §  6,  Th.  1. 

THEOREM  2.  Two  parallel  sections  of  a  hyperboloid  or  a  cone 
are  conies  of  the  same  type.  They  are,  moreover,  similar  and 
similarly  placed,  or,  in  the  case  of  two  hyperbolas,  each  is  similar 
and  similarly  placed  either  to  the  other  or  to  the  conjugate  of  the 
other. 

To  prove  this  theorem,  we  fix  our  attention  on  one  of  the 
surfaces  (1),  say  the  hyperboloid  Hly  and  give  to  k  two  arbi- 
trarily chosen  values,  ki  and  k2,  thus  obtaining  two  arbitrary 
'parallel  sections  of  HI.  The  coefficients  of  the  quadratic 
terms  in  the  two  equations  (3)  which  result  are  respectively 
equal,  since  these  coefficients  in  the  general  equation  (3)  do 
not  contain  k.  Hence,  by  p.  260,  Ex.  40,  the  directrices  of  the 
cylinders  defined  by  the  two  equations  are  similar-  and  simi- 
larly placed  conies,  or,  in  the  case  of  two  hyperbolas,  each  is 
similar  or  similarly  placed  either  to  the  other  or  to  the  conju- 
gate of  the  other.  Consequently,  by  Ch.  XXII,  §  6,  Th.  2, 


QUADRIC   SURFACES  563 

this  is  true  also  of  the  sections  of  the  cylinders  by  the  two 
planes,  q.  e.  d. 

THEOREM  3.  The  sections  of  two  conjugate  hyperboloids  and 
the  common  asymptotic  cone  by  the  same  plane  or  by  parallel 
planes  are  similar  and  similarly  placed  conies,  or,  in  the  case  of 
hyperbolas,  one  of  any  two  is  similar  or  similarly  placed  either 
to  the  other  or  to  the  conjugate  of  the  other. 

It  is  sufficient  to  prove  the  theorem  for  the  sections  of  the 
three  surfaces  by  a  single  plane,  since  its  truth  for  sections  by 
parallel  planes  will  then  follow  from  Theorem  2.  Here,  then, 
k  is  fixed,  and  A  takes  on  successively  the  values  1,  —  1,  0. 
But  the  coefficients  of  the  quadratic  terms  in  (3)  do  not 
contain  X,  and  hence  we  reach  the  desired  conclusion  imme- 
diately, by  reasoning  identical  with  that  used  in  the  proof  of 
Theorem  2. 

The  following  theorem  is  now  obvious. 

THEOREM  4.  A  plane  which  intersects  a  hyperboloid  or  a  cone, 
but  not  in  a  non-degenerate  conic,  cuts  it  in  a  degenerate  conic, 
which  is  of  the  same  type  as  any  section  by  a  parallel  plane. 

Accordingly,  to  ascertain  the  type  of  conic  (degenerate  or 
non-degenerate)  in  which  a  plane  intersects  a  hyperboloid,  it 
is  necessary  merely  to  determine  the  type  of  degenerate  conic 
in  which  the  parallel  plane  through  the  center  meets  the 
asymptotic  cone.  But  the  planes  through  the  center  intersect 
the  cone  in  degenerate  conies  of  all  three  types.  Consequently, 
a  hyperboloid  has  sections  of  all  three  types. 

EXERCISES 

1.  Show  that  every  plane  section  of  an  ellipsoid  is  an  ellipse 
and  that  parallel  sections  are   similar   and   similarly   placed 
ellipses. 

2.  Prove  that  an  elliptic  paraboloid  has  no  hyperbolic  sec- 
tions, that  sections  by  parallel  planes  cutting  the   axis   are 
similar   and   similarly  placed   ellipses,  and  that  sections   by 


564  ANALYTIC   GEOMETRY 

parallel  planes  parallel  to  the  axis  are   equal  and   similarly 
placed  parabolas. 

3.  Prove  that  a  hyperbolic  paraboloid  has  no  elliptic  sec- 
tions, that  sections  by  two  parallel  planes  cutting  the  axes 
are  hyperbolas,  one  of  which  is  similar  and  similarly  placed 
either  to  the  other  or  to  the  conjugate  of  the  other,  and  that 
sections  by  parallel  planes  parallel  to  the  axis,  but  not  to.  a 
directrix  plane,  are  equal  and  similarly  placed  parabolas. 

6.   Circular  Sections.     Consider  a  section  of  the  ellipsoid 


by  a  plane  M  passing  through  the  axis  of  y.  The  section  is  an 
ellipse,  one  of  whose  axes  is  always  the  mean  axis,  26,  of  the 
ellipsoid,  no  matter  how  M  is  situ- 
ated.  When  M,  starting  from  the 
(x,  2/)-plane,  rotates  in  either  direc- 
tion into  coincidence  with  the  (y,  z)- 
plane,  the  second  axis  of  the  ellipse, 
starting  from  the  major  axis,  2  a,  of 
the  ellipsoid,  decreases  continuously 
to  the  minor  axis,  2c.  Consequently, 
there  must  be  a  single  position  of  M, 
in  each  direction  of  rotation,  for 

which  the  second  axis  of  the  ellipse  takes   on  the  value  2b 
equal  to  the  first.     But  then  the  ellipse  is  a  circle. 

These  two  positions,  KOB  and  LOB,  of  the  plane  M  can 
be  constructed  by  describing  in  the  upper  half  of  the  (z,  x)- 
plane  a  semicircle  whose  center  is  at  0  and  whose  radius  is  &. 
The  semicircle  will  meet  the  ellipsoid  in  the  desired  points  K 
and  L. 

Since  the  sections  of  (1)  by  the  planes  KOB  and  LOB  are 
circles,  so  also  are  the  sections  by  planes  parallel  to  KOB  and 
LOB,  by  §  5,  Ex.  1.  The  ellipsoid  has,  then,  these  two  sets  of 
circular  sections  and,  as  can  be  shown  (Ex.  1),  only  these  two. 


QUADRIC   SURFACES  565 

It  can  be  proved  in  the  same  way  that  the  hyperboloid  of 
one  sheet 


contains  just  two  sets  of  circular  sections.  Hence  it  follows, 
by  §  5,  Th.  3,  that  this  is  true  also  of  the  cone  and  the  hyper- 
boloid of  two  sheets.  It  is  to  be  noted,  however,  that  there 
are  no  circular  sections  of  the  cone  by  planes  through  the 
vertex  and  none  of  the  hyperboloid  of  two  sheets  by  planes 
through  the  center. 

The  results  obtained  we  now  consolidate  into  a  theorem.  , 

THEOREM.  An  ellipsoid,  a  hyperboloid,  or  a  cone,  which  is 
not  a  surface  of  revolution,  contains  just  two  sets  of  circular 
sections. 

If,  in  Fig.  9,  b  approaches  a  as  its  limit,  the  planes  KOB 
and  LOB  both  approach  as  their  limits  the  (x,  y)-plane.  Con- 
sequently, an  ellipsoid  of  revolution  has  but  one  set  of  circu- 
lar sections.  This  is  true  also  of  the  hyperboloids  and  cones 
of  revolution. 

The  circles  in  which  the  planes  KOB,  LOB  intersect  the 
ellipsoid  evidently  lie  on  the  sphere  whose  center  is  at  0  and 
whose  radius  is  b  : 

(3)  *l2  +  ^  +  -  =  l. 

62      62      62 

They  therefore  lie  on  the  surface  whose  equation  results  from 
subtracting  (1)  from  (3)  : 


or  c2(a2-6-)»2-a2(62-c2)22 

But  this  surface  consists  of  the  two  planes 


(4)  c  Va2  -  62  x  ±  a  V&2  -  c2  z  =  0. 

Consequently,  these  are   the  equations  of   the   planes  KOB, 
LOB. 


566  ANALYTIC   GEOMETRY 


EXERCISES 

1.  To  show  that  the  ellipsoid  (1)  has  but  two  sets  of  circu- 
lar sections,  prove  first,  using  the  fact  that  the  centers  of  the 
circles  of  any  set  lie  on  a  line  (§  8,  Problem  1),  that  every 
circular  section  must  be  symmetric  in  a  principal  plane ;  then 
show  that  a  section  of  (1)  by  a  plane  passing  through  the  x- 
axis  or  the  z-axis  is  never  a  circle. 

2.  Prove   geometrically  that   the  hyperboloid  (2)  has  two 
sets  of  circular  sections.     Give  a  construction  for  the  planes 
through  the  origin  which  yield  circular  sections. 

3.  Find  the  equations  of  the  planes  just  mentioned. 

4.  Show  that  the  elliptic  paraboloid  (1),  §  (3),  where  a  >  6, 
has  two  sets  of  circular  sections,  by  proving  first  that  this  is 
true  of  the  elliptic  cylinder 

C2          tft 

5+fc-i.          ' 

5.  A    hyperbolic    paraboloid    has    no    circular    sections. 
Why? 

7.  Tangent  Lines  and  Planes.  Let  the  line  L  through  the 
point  PQ  :  (x0,  y0,  z0)  with  the  direction  cosines  cos  a  =  A, 
cos  ft  =  PL,  cos  y  =  v  meet  the  ellipsoid 

(1)  £  +  £  +  ^1 
a?^b*     c* 

in  two  distinct  points,  Pt  and  P2.     To  find  the  coordinates, 
(xi,  y\>  zi)  and  (X2,  K2>  «2)j  of  Pl  and  P2. 
The  parametric  representation  of  L  is 

(2)  x  =  x0  +  Xr,        y  =  y0  +  p.r,        z  =  z0  +  vr, 

where  r  is  the  algebraic  distance  from  P0  to  P :  (x,  y,  z) ;  cf . 
Ch.  XX,  §  8.  The  point  P  of  L  lies  on  the  ellipsoid,  if  and 
only  if  its  coordinates  (x,  y,  z),  as  given  by  (2),  satisfy  (1), 
that  is,  if  and  only  if  r  satisfies  the  equation 


QUADRIC   SURFACES  567 


Since  L  intersects  the  ellipsoid  in  two  distinct  points,  (3) 
has  two  distinct  roots.  If  we  denote  them  by  r}  and  r2,  the 
coordinates  of  P±  and  P2  are 

yl  =  t/0  +  /An,        Zi  =  «0  +  ^'i  5 


Tangent  Line.  Suppose,  now,  that  P0  lies  on  the  ellipsoid. 
Then  one  of  the  points  of  intersection  of  L,  say  P2,  coincides 
with  PQ.  Analytically,  we  have 

ff  2          ,,2          M  Z 

(A\  ^0    _|_  yp,     I     Z0    —  1 

a2+62+^- 

so  that  the  absolute  term  in  (3)  is  zero  ;  also,  rz  =  0  and  r!  is 
the  distance  from  P0  to  Px. 

Imagine  a  curve  drawn  on  the  surface  through  P0  and  Pt, 
for  example,  an  arbitrary  plane  section,  C,  through  P0  and  Pj. 
The  line  L  is  the  secant  P0Pi  of  C  and  its 
limiting  position,  as  Pl  moving  along  C 
approaches  P0  as  its  limit,  is  the  tangent 
to  C  at  P0.  We  define  this  tangent  as  the 
tangent  line  to  the  surface  at  P0  in  the  direc- 
tion of  the  curve  C.  Fia.  10 

When  P  approaches  P0,  then  no  matter 

what  curve  C  of  approach  is  chosen  rj  approaches  zero.  But, 
when  r±  approaches  zero,  the  coefficient  of  r  in  (3)  approaches 
zero,  and  conversely.  Consequently,  the  line  L  is  a  tangent  to 
the  ellipsoid  (1)  at  the  point  P0  on  the  ellipsoid  if  and  only  if 

(5}  g(A  .  yw  .  V-Q 

a2        &2       c2 

Tangent  Plane.  There  are  evidently  infinitely  many  lines 
L  tangent  to  the  surface  at  P0.  For  them  X,  p,  v  have  varying 


568  ANALYTIC   GEOMETRY 

values,  which,  however,  always  satisfy  (5).  To  obtain  the 
locus  of  all  the  tangent  lines  L,  we  have  only  to  eliminate  the 
auxiliary  variables  A,  p,  v,  r  from  equations  (2)  and  (5).  Sub- 
stituting the  values  of  X,  p.,  v  as  given  by  (2)  into  (5)  and  sup- 
pressing the  factor  1/r,  we  get  the  equation 


o   | 


which  reduces,  by  virtue  of  (4),  to 

/6\  £%*  ,  M  ,  «o?  _  i 

a*  T  6*      c2 

But  this  is  the  equation  of  a  plane.  Hence  we  have  the 
theorem  : 

THEOREM  1.  The  tangent  lines  at  a  point  PQ  of  an  ellipsoid 
all  lie  in  a  plane. 

The  plane  of  the  tangent  lines  at  P0  we  define  as  the  tan- 
gent plane  to  the  ellipsoid  at  P0.  Its  equation  is  given  by  (6). 

EXERCISES 

Find  for  each  of  the  following  surfaces  the  condition  that 
a  line  L  through  a  point  P0  of  the  surface  be  tangent  to  the 
surface.  Prove  the  analogue  of  Theorem  1  and  deduce  the 
equation  of  the  tangent  plane  at  P0. 

1.  The  unparted  hyperboloid.     2.  The  biparted  hyperboloid. 

3.  The  elliptic  paraboloid.         4.  The  hyperbolic  paraboloid. 

5.  The  cone,  P0  not  being  at  the  vertex. 

6.  Prove  that  the  tangent  plane  to  a  hyperboloid  of   one 
sheet  at  a   point  P0  is  the  plane  determined  by  the  rulings 
which  pass   through  P0.     Hence  show  that   the  two   sets  of 
rulings   found   in  §  4   exhaust  all   the  straight   lines  on  the 
surface. 

7.  The  same  for  a  hyperbolic  paraboloid. 

8.  Let    Q  be  a  quadric  surface,  not  a  cone  or  a  cylinder. 
Prove  that  a  plane  is  tangent  to  Q  if  and  only  if  it  intersects 


QUADRIC   SUKFACES  569 

Q  in  a  point  (a  degenerate  ellipse)  or  in  two  intersecting  lines 
(a  degenerate  hyperbola). 

8.  Diameters.  Diametral  Planes.  Problem  1.  Find  the 
locus  of  the  centers  of  parallel  sections  of  the  ellipsoid 

999 
T**  1  /*  *y* 

(!)  L2  +  r2  +  l  =  1- 

a2      b2      <? 

Let  the  common  normals  to  the  planes  of  the  sections  have 
the  direction  components  A,  B,  C.  Let  P:  (X,  T,  Z)  be  the 
center  of  one  of  the  sections  and  let  L,  with  the  direction 
cosines  X,  p.,  v,  be  an  arbitrary  line  through  P,  which  lies  in 
the  plane  of  this  section.  The  parametric  equations  of  L  are, 
then, 
(2)  x  =  X  +  \r,  y=Y+fir,  z=Z  +  vr, 

where  r  is  the  algebraic  distance  from  P  to  (x,  y,  z). 

Since  the  points  of  intersection  of  L  with  the  ellipsoid  are 
equally  distant  from  P,  their  algebraic  distances,  r±  and  r2, 
from  P  are  negatives  of  each  other  :  rx  -f  r»  =0.  But  rl}  r2  are 
the  roots  of  the  quadratic  equation 


and  consequently,  by  Ch.  XIII,  §  5, 


This  equation  says  that  the  direction  whose  components  are 
X/a?,  Y/&,  Z/c2  is  always  perpendicular  to  L.  But  L  is  an 
arbitrary  line  in  one  of  the  planes  of  the  sections  and  the  only 
direction  which  is  always  perpendicular  to  it  is  that  of  the 
normals  to  these  planes,  that  is,  the  direction  whose  compo- 
nents are  A,  B,  C.  Consequently,  X/cf-,  Y/b"1,  Z/c2  are  pro- 
portional to  A,  B,  C,  or 

(5)  A  =  ^.  =  _^. 

a2  A     WB     <?C 


570  ANALYTIC   GEOMETRY 

The  centers  of  the  sections  lie,  then,  on  a  line  through  the 
center  of  the  ellipsoid.  Such  a*  line  is  known  as  a  diameter. 
Accordingly,  we  can  state  our  result  as  follows. 

THEOREM  1.  The  locus  of  the  centers  of  parallel  sections  of 
the  ellipsoid  (1)  is  that  portion  of  a  diameter  which  lies  within 
the  ellipsoid.  If  the  direction  components  of  the  normals  to  the 
planes  of  the  sections  are  A,  B,  C,  those  of  the  diameter  are  a2  A, 


Exercise.  The  tangent  planes  at  the  extremities  of  a 
diameter  are  parallel  to  the  sections  whose  centers  the  diame- 
ter contains. 

Problem  2.  Find  the  locus  of  the  mid-points  of  a  set  of 
parallel  chords  of  the  ellipsoid  (1). 

Let  X,  p.,  v  be  the  given  direction  cosines  of  one  of  the  chords 
and  let  P  :  (X,  Y,  Z)  be  the  mid-point  of  the  chord.  Then 
equations  (2)  represent  the  chord  parametrically. 

Since  P  is  the  mid-point  of  the  chord,  the  algebraic  dis- 
tances, ri  and  r«,  from  it  to  the  end  points  of  the  chord  are 
negatives  of  each  other  :  rv  +  r2  =  0.  Hence,  as  in  Problem  1, 
we  obtain  the  equation  (4).  But,  whereas  in  that  problem 
X,  /*,  v  were  auxiliary  variables,  here  they  are  given  constants, 
and  hence  (4)  is  the  equation  satisfied  -  by  the  point  P  of  the 
locus.  But  (4)  represents  a  plane  through  the  center  of  the 
ellipsoid.  Such  a  plane  is  known  as  a  diametral  plane.  Thus 
our  result  is  : 

THEOREM  2.  The  locus  of  the  mid-points  of  a  set  of 
parallel  chords  of  the  ellipsoid  (1)  is  that  portion  of  a  dia- 
metral plane  lying  within  the  ellipsoid.  If  the  direction  com- 
ponents of  the  chords  are  I,  m,  n,  the  equation  of  the  diametral 
plane  is 


-2 

c2 


QUADRIC  SURFACES  571 

Conjugate  Diameters  and  Diametral  Planes. 

THEOREM  3.  If  a  diameter  D  contains  the  centers  of  sections 
parallel  to  a  diametral  plane  M,  then  M  bisects  the  chords 
parallel  to  D,  and  conversely. 

For,  let  D  have  the  direction  com- 
ponents I,  m,  n,  and  let  M  be  the  plane 

Ax  +  By  +  Cz  =  0. 

The  condition  that  D  contain  the  cen- 
ters of  sections  parallel  to  M  is,  by  Th. 
1,  that  %  '  FIG.  11 

l:m:n  =  a2A:b2£:  c2C. 

The  condition  that  M  bisect  the  chords  parallel  to  D  is,  by 
Th.  2,  that 

4.<*:CN.ifS:£. 

a2  62  c2 

The  two  conditions  can  both  be  written  in  the  form 

(K\  ?          m        n 

~a*A~WB~&C' 

% 

and  are,  therefore,  equivalent,  q.  e  .d. 

A  diameter  D  and  a  diametral  plane  M  in  the  relationship 
described  are  said  to  be  conjugate.  We  have,  then,  the  follow- 
ing theorem. 

THEOREM  4.  The  diameter  D  with  the  direction  components 
I,  m,  n  and  the  diametral  plane  Ax  +  By  +  Cz  =  0  are  conjugate 
if  and  only  if 

fR\  I      _  m        n 

a*A~tfB~~c*C' 

Exercise.  Show  that  an  axis  and  the  principal  plane  per- 
pendicular to  it  are  conjugate,  and  that  in  no  other  case  is  D 
perpendicular  to  its  conjugate,  M. 

THEOREM  5.  If  two  diameters,  DI  and  D2,  are  conjugate  in 
the  ellipse  E  in  which  their  plane  meets  the  ellipsoid,  each  lies  in 
the  diametral  plane  of  the  other. 


572 


ANALYTIC   GEOMETRY 


E 


FIG.  12 


For,  since  Dt  is  conjugate  to  D2  in  E,  DI  bisects  the  chords 
of  E  parallel  to  D2.  But  the  diametral  plane  M2  conjugate  to 
D2  also  bisects  these  chords.  Hence 
DI  must  be  the  line  in  which  M2  meets 
the  plane  of  E,  and  so  DI  lies  in  M2. 
Similarly,  D2  lies  in  the  diametral  plane, 
M1}  conjugate  to  Dt. 

THEOREM  6.  If  one  diameter  lies  in 
the  diametral  plane  conjugate  to  a  second, 
then  the  second  diameter  lies  in  the  diametral  plane  conjugate  to 
the  first. 

Suppose  that  D±  lies  in  the  diametral  plane  Mz  conjugate  to 
D2.  It  will  follow,  then,  by  Th.  5,  that  D2  lies  in  the  di- 
ametral plane  Ml  conjugate  to  D1}  if  we  can  show  that  Dl  and 
D2  are  conjugate  diameters  in  the  ellipse  E  (Fig.  12).  This  is 
the  case,  for,  since  M2  bisects  all  chords  parallel  to  D2,  then 
D!  bisects  all  chords  of  E  parallel  to  D2. 

Conjugate  Diameters.  Conjugate  Diametral  Planes.  Given 
three  diameters  DI,  D2,  D3  and  three  diametral  planes  MI,  M2, 
Mz  such  that  DI,  D?,  D3  are  the  lines  of 
intersection  of  M1}  -M2,  M3  or  Ml}  M.2,  M3 
are  the  planes  determined  by  DI,  D2,  D3 
(Fig.  13).  Consider  the  following  re- 
lationships : 

RI  :   DI,  D2,  D3  and  Mlt  M2,  M3  are  re- 
spectively conjugate ; 
Each  diameter  contains  the  centers  of  sections  parallel  to 

the  plane  of  the  other  two  ; 
Each  diametral  plane  bisects  the  chords  parallel  to  the 

line  of  intersection  of  the  other  two. 

According  to  the  definition  of  conjugacy  of  a  diameter  and  a 
diametral  plane,  these  relationships  are  equivalent : 

THEOREM  7.  Any  one  of  the  relationships  R  is  equivalent  to 
each  of  the  other  two ;  that  is,  if  any  one  holds,  so  does  each  of 
the  other  two. 


FIG.  13 


R 


R. 


QUADRIC   SURFACES  .       573 

Three  diameters  in  the  relationship  R2  are  called  conjugate 
diameters,  and  three  diametral  planes  in  the  relationship  R3 
are  called  conjugate  diametral  planes. 

There  are  infinitely  many  sets  of  three  diameters  and  three 
diametral  planes  in  the  relationship  Rv.  For,  let  E  be  an 
arbitrary  section  of  the  ellipsoid  by  a  plane  through  0,  and 
let  Dl  and  D2  be  any  two  diameters  conjugate  in  E.  Then 
the  diametral  planes  MI  and  M2  conjugate  to  D±  and  D2  will, 
by  Th.  5,  pass  through  D2  and  Dt  respectively.  Finally,  the 
diametral  plane,  M3)  conjugate  to  the  diameter,  D3,  in  which 
MI  and  M2  intersect  will,  by  Th.  6,  contain  DI  and  D2  and 
hence  must  be  the  plane  of  DI  and  D2. 

The  following  theorems  follow  directly  by  application  of 
Theorems  2  and  1,  respectively. 

THEOREM  8.  The  diameters  with  the  direction  components 
li}  mi,  ni}  12,  m2,  n2,  Z3,  m3,  n3  are  conjugate  if  and  only  if 


,2         t2  ,        2  _  ft 

-    —  j—  -  •  —  j—  -  —   \)  * 

a2         6-          c2 


m2m3        23  _  A 
"-:()> 


, 
^"H 

THEOREM  9.     The  diametral  planes 

Ajx  +B&  +  C}z  =  0,  A2x  +  B2y  +  C&  =  0,  A3x  +  B3y  +C3z  =  0 
are  conjugate  if  and  only  if 

tfAiAt  +  WBiBz  +  c-Ci(72  =  0, 
a2^2^3  +  Z>2B2B3  +  C2(72(73  =  0, 
a*A3Ai  +  VB.^  +  c^CsC,  =  0. 

EXERCISES 

State  and  prove  for  the   following  surfaces    the   theorems 
analogous  to  Theorems  1,  2. 
1.    The  unparted  hyperboloid. 


574  ANALYTIC  GEOMETRY 

2.  The  biparted  hyperboloid. 

3.  The  elliptic  paraboloid.     Show,  in  particular,  that  the 
diameters  and  diametral  planes  are  all  parallel  to  the  axis. 

4.  The  hyperbolic  paraboloid.     Describe  the  positions  of 
the  diameters  and  diametral  planes. 

Give  reasons  for  the  following  exceptions  to  the  theorems 
just  proved. 

5.  Hyperboloids.     There  is  no  analogue  to  Theorem  1  for 
parallel  sections  parallel  to  an  element  of  the  asymptotic  cone, 
and  no  analogue  to  Theorem  2  for  lines  parallel  to  an  element 
of  the  asymptotic  cone. 

6.  Paraboloids.     There  is  no  analogue  to  Theorem  1  for 
parallel  sections  parallel  to  the  axis,  and  no  analogue  to  Theo- 
rem 2  for  lines  parallel  to  the  axis.     In  the  case  of  a  hyper- 
bolic paraboloid  there  is  also  no  analogue  to  Theorem  2  for  a 
set  of  parallel  lines  parallel  to  a  directrix  plane. 

State  and  prove  for  the  following  surfaces  the  theorems 
analogous  to  Theorems  3,  4. 

7.  The  imparted  hyperboloid. 

8.  The  biparted  hyperboloid. 

9.  There  are  no  conjugate  diameters  and  diametral  planes 
for  a  paraboloid.     Why  ? 

10.  Find  for  the  ellipsoid  (1)  the  equation  of  the  diametral 
plane  conjugate  to  the  diameter  through  the  point  (x0)  y0,  z0) 
of  the  surface. 

11.  Prove  that  the  pairs  of  conjugate  diameters  and  dia- 
metral planes  are  the  same  for  two  conjugate  hyperboloids. 

12.  Discuss  the  conjugacy  of  three  diameters  or  three  dia- 
metral planes  for  either  hyperboloid. 

13.  Prove  Theorems  8,  9. 

14.  State  and  prove  the  analogues  of  Theorems  8,  9  for 
either  hyperboloid. 


QUADRIC   SURFACES  575 

9.   Poles  and  Polars.     Through  a  point  PQ  :  (x0,  y0,  z0)  not  on 
the  ellipsoid  : 


an  arbitrary  line  L  is  drawn  meeting  the  ellipsoid  in  Qt  and 
Q2-  What  is  the  locus  of  the  point  P  which  with  P0  divides 
QiQ2  harmonically? 

By  the  method  used  in  solving  the  corresponding  problem 
in  the  plane,  Ch.  XIV,  §  9,  the  locus  is  found  to  be  the  plane 


or  a  portion  of  this  plane. 

The  point  P0  (not  on  the  ellipsoid)  and  the  plane  (2)  are 
said  to  be  pole  and  polar  in  the  ellipsoid  :  P0  is  the  pole  of  (2), 
and  (2)  the  polar  of  P0-  A  point  on  the  ellipsoid  and  the 
tangent  plane  at  the  point  are  defined  to  be  pole  and  polar. 

By  the  methods  of  Ch.  XIV,  §§  9-11,  the  following  theorems 
can  now  be  proved. 

THEOREM  1.  Let  Q  be  a  central  quadric  (an  ellipsoid  or 
hyperboloid).  Every  point  in  space,  except  the  center  of  Q,  has 
a  polar  with  respect  to  Q. 

THEOREM  2.  Every  point  in  space  has  a  polar  with  respect 
to  a  paraboloid. 

THEOREM  3.  Let  Q,  be  a  central  quadric  or  a  paraboloid. 
Every  plane  in  space,  which  is  not  a  diametral  plane  of  Q,  has 
a  pole  with  respect  to  Q. 

The  poles  and  polars  considered  in  the  following  theorems 
are  taken  with  respect  to  an  arbitrarily  chosen  central  quadric 
or  paraboloid. 

THEOREM  4a.  If  one  point  lies  in  the  polar  plane  of  a  second, 
the  second  point  lies  in  the  polar  plane  of  the  first. 

THEOREM  46.  If  one  plane  contains  the  pole  of  a  second,  the 
second  plane  contains  the  pole  of  the  first. 


576  ANALYTIC   GEOMETRY 

THEOREM  5a.     If  a  number  of  points  lie  on  a  line  L,  their 
polar  planes  pass  through  a  line  L'  or  are  parallel. 

L]  THEOREM  56.  If  a  number  of  planes  pass 
through  a  line  L'  (or  are  parallel),  their  poles 
lie  on  a  line  L. 


Theorems  5a,  56  are  peculiar  to  the  geometry 
of  space.     We  give  a  proof  of  Theorem  5a. 

Let  P^fo,  yl}  Zi),  P2:(«2,   2/2,  z2)   be.  two 
FIG.  14          distinct  points.     Their  polar  planes  are 

WEE^  +  M  +  ^_i=0,        V^  +  M  +  ^_1  =  0. 
or       cr       cr  a2        62        c2 

Let  P3 :  (xs,  2/3,  z3)  be  an   arbitrary  point  of  the   line 
Then,  by  Ch.  XXI,  §  2, 

x3  =  pxi  +  (1  -  p)xt,  2/3  =  pyi  +  (1  -  p)2/2,  «3  =  pzi  +  (1  - 
Consequently,  the  polar  plane  of  P3,  namely 

XsX.ysV  ,  z3z      -,  _ft 

— i~  """  TF   "~  ~Ia"  ~  ' 

a2       62       c2 

can  have  its  equation  written  in  the  form 
pu  +  (1  —  p)  v  =  0, 

and  hence  passes  through  the  line  of  intersection  of  the  polar 
planes  of  Pt  and  P2,  if  they  intersect,  or  is  parallel  to  them,  if 
they  are  parallel,  q.  e.  d. 

Two  lines  L  and  L',  in  the  relationship  described  in  The- 
orem 5a  or  56,  are  each  said  to  be  polar  or  conjugate  to  the 
other.  The  following  theorems  concerning  polar  lines  are 
readily  proved ;  cf.  Ths.  8a,  86  of  Ch.  XIV,  §  11. 

THEOREM  6a.  The  polar  of  a  line  intersecting  the  quadric  in 
two  distinct  points  is  the  line  of  intersection  of  the  tangent  planes 
at  these  points. 

THEOREM  66.  The  polar  of  a  line  not  meeting  the  quadric  is 
the  line  joining  the  points  of  contact  of  the  two  planes  through  the 
line  tangent  to  the  quadric. 


QUADRIC   SURFACES  577 

THEOREM  6c.  The  polar  of  a  tangent  to  a  quadric,  not  a 
ruling,  is  a  second  tangent  line  with  the  same  point  of  contact. 
A  ruling  of  a  quadric  is  self-polar. 

EXERCISES 

Establish  formula  (2)  and  the  analogous  formulas  for  the 
other  quadrics.  Prove  the  theorems  stated  without  proof  in 
the  text. 

Discuss  poles  and  polars  with  respect  to  a  sphere  (cf.  Ch. 
XIV,  §  9,  Exs.  9-11),  showing,  in  particular,  that  two  polar 
lines  are  always  perpendicular. 

10.  One-Dimensional  Strains,  with  Applications.*  The  one- 
dimensional  strain  which  stretches  all  space  directly  away  from 
the  (y,  z)-plane  (or  compresses  all  space  directly  towards  the 
(y,  z)-plane),  so  that  each  point  is  carried,  along  a  parallel  to 
the  axis  of  x,  to  a  times  its  original  distance  from  the  (?/,  2)- 
plane,  where  a  is  a  positive  constant  not  unity,  has  the  equa- 
tions 
i)  x'  =  ax,  y'  =  y,  z'  =  z. 

Similarly,  the  equations 

ii)  x'  =  x,        y'  =  by,        z'  =  z, 

iii)  x'  =  x,       y'  =  y,         z'  =  cz, 

where  b  and  c  are  positive  constants  different  from  unity, 
represent  one-dimensional  strains  in  the  directions  of  the  axes 
of  y  and  z  respectively. 

One-dimensional  strains  have  the  following  properties  : 

A.  Planes  go  into  planes,  and  hence  straight  lines  go  into 
straight  lines ; 

B.  Parallel  planes  go  into  parallel  planes  and  hence  paral- 
lel straight  lines  go  into  parallel  straight  lines  ; 

C.  Tangent  surfaces  go  into  tangent  surfaces,  and  tangent 
curves  into  tangent  curves. 

*  Cf.  Ch.  XIV,  particularly  §  5. 


578  ANALYTIC   GEOMETRY 

One-dimensional  strains  do  not  in  general  preserve  angles 
or  areas.  They  never  preserve  volumes  ;  for  example,  i)  car- 
ries a  portion  of  space  of  volume  V  into  a  portion  of  space  of 
volume  aV. 

The  product  of  the  three  one-dimensional  strains  i),  ii),  iii) 
is  the  transformation 

T:  x'  =  ax,        y'  =  by,        zr  =  cz. 

It  is  clear  from  the  foregoing  that  T  carries  a  portion  of  space 
of  volume  V  into  a  portion  of  space  of  volume  abc  V. 

Applications.     The  sphere 
(1)  x*  +  y*  +  z*  =  l 

is  carried  by  the  transformation  T  into  the  ellipsoid 


The  volume  of  the  sphere  is  |TT.     That  of  the  ellipsoid  is, 
then, 


THEOREM  1.     The  volume  of  the  ellipsoid  (2)  is 

V=  f  irabc. 
Let  the  triples 

(3)  AX,  /*!,  vi,         A2,  pv,  v2,         A3,  p.3,  v3 

be  the  direction  cosines  of  three  mutually  perpendicular  (and 
hence  conjugate)  diameters,  Di}  D2,  D3,  of  the  sphere  (1). 
They  are,  then,  also  the  coordinates  of  three  points,  Plt  P2,  Pa, 
on  the  sphere,  which  are  respectively  extremities  of  Dj,  DZ)  D3. 
Now  T  carries  Dlt  D2,  D3  into  three  diameters,  ZV>  D-t,  D3't 
of  the  ellipsoid  (2),  and  carries  PI,  P2,  P3  into  three  points, 
PI,  PZ,  PS'>  on  the  ellipsoid,  which  are  respectively  extremi- 
ties of  ZV,  D2',  D3'.  Evidently,  the  triples 

(4)  aAi,  6/*i,  cvj,         aAa,  b^,  cv2,        aA3,  bp*,  Cv3 

are  both  the  coordinates  of  P/,  P2'»  PS'  an{l  the  direction  com- 
ponents of  DI,  D2',  D3'. 


QUADRIC   SURFACES  579 

Considered  as  the  direction  components  of  A'>  A'>  A'>  the 
triples  (4)  satisfy  the  conditions  of  Th.  8,  §  8.  Hence  we 
have  the  theorem : 

THEOREM  2.  The  transformation  T  carries  three  mutually 
perpendicular  (and  therefore  conjugate)  diameters  of  the  sphere 
(1)  into  three  conjugate  diameters  of  the  ellipsoid  (2). 

It  follows,  by  Th.  7,  §  8,  that  T  carries  three  mutually 
perpendicular  diametral  planes  of  (1)  into  three  conjugate 
diametral  planes  of  (2),  and  carries  a  diameter  and  the  per- 
pendicular diametral  plane  of  (1)  into  a  diameter  and  the  con- 
jugate diametral  plane  of  (2). 

Considered  as  the  coordinates  of  P/,  P2',  P3',  the  triples 
(4)  give  immediately,  as  the  squares  of  the  half-lengths  of  the 
diameters  A'>  A'>  A'> 

aV  +  &  V  +  cV,     aV  +  &  V  +  cV,    «2A32  +  &W  +  cV- 
The  sum  of  these  squares  is 

a2(Aj2  +  V  +  A32)  +  b*-(tf  +  tf  +  M32)+  c>t2  +  v22  +  v32). 

Since  the  triples  (3)  are  the  direction  cosines  of  three  mutu- 
ally perpendicular  lines,  so  also  are  the  triples 

A!,  AS,  A3,         pi,  /x2,  fj.3,         vi,  v2,  v3 ; 
cf.  Ch.  XXIV,  §  6.     Hence  the  above  sum  has  the  value 

tt2  +  &2'+  C2 

and  is  therefore  independent  of  the  three  conjugate  diameters 
A'i  A',  A'  taken. 

THEOREM  3.  The  sum  of  the  squares  of  the  lengths  of  three 
conjugate  diameters  of  an  ellipsoid  is  constant. 

EXERCISES 

1.  Prove  analytically  the  properties  A,  B,  C  of  one-dimen- 
sional strains. 

2.  What  angles  and  what  areas  does  the  transformation  i) 
preserve  ? 


580  ANALYTIC   GEOMETRY 

3.  Prove  that  i)  carries  a  region  of  volume  V  into  a  region 
of  volume  a  V. 

4.  Show  that  T  carries  a  line  with  the  direction  components 
Z,  m,  n  into  a  line  with  the  direction  components  al,  bm,  en. 

5.  The  plane  M  goes  into  the  plane  M'  under  T.     If  the 
direction  components  of  the  normals  to  M  are  A,  B,  C,  what 
are  those  of  the  normals  to  M'  ? 

6.  Assuming  the  equation  of  the  tangent  plane  to  the  sphere 
(1)  at  the  point  (x0,  y0,  z0),  deduce  by  means  of  the  transforma- 
tion T  the  equation  of  the  tangent  plane  to  the  ellipsoid  (2) 
at  the  point  (x0',  y0',  20'). 

7.  Show  that  a  hyperboloid  of  general  type  can  always  be 
carried  into  a  hyperboloid  of  revolution  by  means  of  a  trans- 
formation of  the  form  T. 

EXERCISES  ON   CHAPTER  XXIII 

1.  Find  the  equation  of  the  quadric  surface  generated  by 
the  lines  x  —  Xz  =  0,  \y  —  z  =  0,  where  A  is  a  parameter.     De- 
termine the  equations  of  the  second  set  of  rulings  and  set  up 
a  parametric  representation  of  the  surface. 

2.  The  same  for  the  lines  y  —  A  —  1=0,  \x  —  z  +  2  =  0. 

3.  Find  the  equations  of  the  planes  which  pass  through  the 
line  y  =  2,  x  +  2z  =  0  and  are  tangent  to  the  ellipsoid 

aj«  +  3y2  +  2z2  =  6. 

4.  Prove  that  the  sections  of  the  hyperbolic  paraboloid  and 
hyperbolic  cylinder  : 


by  the  same  plane  or  by  parallel  planes,  oblique  to  the  z-axis, 
are  hyperbolas,  each  of  which  is  similar  and  similarly  placed 
to  the  other  or  to  the  conjugate  of  the  other. 

5.    The  umbilics  of  a  quadric  surface  which  has  circular  sec- 
tions are  the  extremities  of  the  diameters  which  contain  the 


QUADRIC   SURFACES  581 

centers  of  these  sections.     Find  their  coordinates  in  the  case 
of  the  ellipsoid  (1),  §  1. 

SIMILAR  QUADRICS 

Definition.  Two  central  quadrics  are  said  to  be  similar  if 
the  principal  sections  of  one  are  similar,  respectively,  to  the 
principal  sections  of  the  other. 

6.  Prove  that  the  ellipsoids  defined  by  the  equation 

(1)  *+£  +  -*=  *>  X>°> 

a-      b-     c2 

where  A.  is  a  parameter,  are  similar. 

7.  Show   that,    of    the   hyperboloids    represented   by   the 
equation 


those  for  which  A.  is  positive  are  all  similar,  and  that  this  is 
true  also  of  those  for  which  A.  is  negative.  Prove  that  all  the 
hyperboloids  have  the  same  asymptotic  cone. 

8.  Prove  that  all  the  ellipsoids    (1)  have  the  same  pairs 
of  conjugate  diameters  and  diametral  planes. 

9.  The  same  for  the  hyperboloids  (2). 

Definition.  Two  paraboloids  of  the  same  type  are  similar, 
if  the  principal  sections  of  one  are  proportional  in  scale 
(Ch.  VI,  §  1)  to  the  principal  sections  of  the  other. 

10.  Prove  in  each  case  that  the  paraboloids  defined  by  the 
given  equation  are  all  similar  : 


* 

cr      o'  a-      o- 

EULED  SURFACES 

11.  Show  that  the  pencil  of  planes  through  the  ruling  V0  of 
the  hyperboloid  (1),  §  4,  cuts  the  surface  in  the  set  of  rulings 
U  and  that  the  pencil  of  planes  through  C70  cuts  it  in  the 
rulings  V. 


582  ANALYTIC  GEOMETRY 

12.  Let  P  be  a  point  on  the  minimum  ellipse  of  the  hy- 
perboloid  (1),  §  4,  and  let  <£  be  the  common  angle  which  the 
two  rulings  through   P  make  with   the   z-axis.     Prove   that 
tan  <£  =&i/c,  where  &!  is  the  half-length  of  the  diameter  of  the 
minimum  ellipse  which  is  conjugate  to  the  diameter  through  P. 

13.  Using  the  result  of  Ex.  12,  show  that  a  hyperboloid  of 
revolution  of  one  sheet  can  be  generated  by  the  rotation  of  a 
ruling  of  either  set  about  the  axis  which  does  not  meet  the 
surface. 

14.  Prove  that  the  rulings  of  one  set  on  a  hyperbolic  para- 
boloid intercept  proportional  segments  on  two  rulings  of  the 
other  set. 

Loci 

15.  Find  the  locus  of  a  point  which  moves  so  that  its  dis- 
tance from  a  fixed  point   bears  to  its  distance   from  a  fixed 
plane,  not  through  the  point,  a  constant  ratio,  k. 

Ans.  A  quadric  of  revolution  which  is  an  ellipsoid,  an 
elliptic  paraboloid,  or  a  hyperboloid  of  two  sheets,  according 
as  k  is  less  than,  equal  to,  or  greater  than  unity. 

16.  A  point  moves  so  that  its  distance  to  a  fixed  point  bears 
to  its  distance  to  a  fixed  line,  not  through  the  point,  a  constant 
ratio.     Find  its  locus. 

Exercises  17-19.  In  connection  with  these  exercises,  Exs. 
28,  29,  p.  522  will  be  found  useful. 

17.  Find  the  locus  of  a  point  which  moves  so  that  its  dis- 
tances to  two  skew  lines  are  always  in  the  same  ratio,  k. 

18.  Prove  that  a  line  which  is  rotated  about  an  axis  skew 
to  it  generates  a  hyperboloid  of  revolution  of  one  sheet ;  cf. 
Ex.  13. 

19.  Let  L  and  L'  be  two  fixed  skew  lines  and  let  M  and  M ' 
be  two  planes,  which  pass  through  L  and  L'  respectively  and 
so  move  that  they  are  always  mutually  perpendicular.     Find 
the  locus  of  their  line  of  intersection. 


QUADRTC   SURFACES  583 

20.  The  locus  of  a  line  which  so  moves  that  it  always  inter- 
sects three  fixed  skew  lines,  not  parallel  to  a  plane,  is  a  hyper- 
boloid  of  one  sheet.     Prove  this  theorem  in  the  case  that  the 
fixed  lines  are 

f  X  =  C,  f  X  =  —  C,  \X  =  —  Z  COt  0, 

\  y  =  z  cos  0 ;         \  y  =  —  z  cos  6  ;         }  y  =  c  sin  0, 

where  c  =£  0  and  6  =£  0,  n-. 

£ 

21.  The  locus  of  a  line  which  so  moves  that  it  always  inter- 
sects three  fixed  skew  lines,  parallel  to  a  plane,  is  a  hyperbolic 
paraboloid.     Prove  this  theorem  in  the  case  that  one  of  the 
fixed  lines  is  the  axis  of  z  and  the  others  have  the  equations 
x  =  c,  z  =  my ;  x  =  —  c,  z  =  —  my,  where  cm  3=  0. 

22.  A  line  moving  so  that  it  is  always  parallel  to  a  fixed 
plane,  M,  and  always  intersects  two  fixed  skew  lines,  neither 
of  which  is  parallel  to  M,  generates  a  hyperbolic  paraboloid. 
Prove  this  theorem  when  M  is  the  (x,  z)-plane  and  the  two  fixed 
lines  are  the  last  two  of  the  three  in  Ex.  21. 


CHAPTER   XXIV 


SPHERICAL  AND  CYLINDRICAL   COORDINATES. 
FORMATION  OF  COORDINATES 


TRANS- 


1.  Spherical  Coordinates.  Given  a  point  0,  a  ray  OA  issu- 
ing from  0,  and  a  half-plane  m  bounded  by  the  line  of  the  ray 
OA.  Let  P  be  any  point  of  space.  Join  P 
to  O  and  construct  the  half-plane,  #,  deter- 
mined by  OA  and  OP.  Denote  the  distance 
OP  by  r,  the  angle  A  OP  by  <f>,  and  the  angle 
from  the  half-plane  m  to  the  half-plane  p  by  6. 
Then  (r,  <£,  0)  are  the  spherical  coordinates  of 
the  point  P. 

For  a  given  value,  r0,  of  the  radius  vector  r, 
the  point  P  lies  on  a  sphere  whose  center  is 
at  0  and  whose  radius  is  r  =  r0.     The  angle 
6  is  the  longitude  of  P,  measured  from  the 
prime  meridian  m,  and  the  angle  <£  is  the 
colatitude  (complement  of  the  latitude),  at  least  for  a  point  P 
on  the  upper  half  of  the  sphere. 

The  radius  vector  r  is,  by  definition,  positive  or  zero.  The 
colatitude  <f>  shall  be  restricted  to  values  between  0  and  TT  in- 
clusive :  0  <  </>  <  TT.  The  longitude  9  shall  be  unrestricted ;  it 
shall  be  taken  as  positive  if  measured  in  the  direction  shown, 
and  as  negative,  if  measured  in  the  opposite  direction.* 

*  It  is  possible  to  define  spherical  coordinates  so  that  r  or  <j>  or  both 
are  also  unrestricted.  Systems  of  these  extended  types  are  not  oftjen 
necessary,  and  when  exceptional  need  for  them  occurs,  they  can  easily 
be  introduced. 

584 


FIG.  1 


SPHERICAL  AND  CYLINDRICAL  COORDINATES     585 


It  is  clear  that  the  r-  and  ^-coordinates  of  a  given  point  P 
are  unique,  while  the  0-coordinate  has  infinitely  many  values, 
each  two  differing  by  an  integral  multiple  of  2  TT.*  Conversely, 
if  r,  <£,"  e  are  given,  such  that  r  >  0  and  0  <  $  <  IT,  a  unique 
point  P  is  determined. 

Let  r0,  <£0,  00  be  particular  values  of  r,  (f),  e,  such  that  r0  >  0 
and  0  <  <£0  <  TT.  The  equation  r  =  r0  represents  a  sphere, 
whose  center  is  at  0  and  whose  radius  is  r0 ;  <£  =  <j!>0  defines 
one  nappe  of  a  circular  cone  whose  vertex  is  at  0  and.  whose 
axis  lies  along  OA ;  finally  0  =  00  represents  a  meridian  half- 
plane  issuing  from  the  line  of  OA. 

Transformation  to  and  from  Rectangular  Coordinates.  Let 
P  be  any  point  of  space  whose  coordinates,  referred  to  a  system 
of  rectangular  axes,  are  (x,  y,  z).  Let  P  have  the  spherical 
coordinates  (r,  <£,  6)  with  respect  to  0, 
OA,  and  m,  as  chosen  in  the  figure.  It  is 
clear  that 

x  =  ON  cos  e,  y  =  ON  sin  0, 

and     ON  =  r  sin  <f>,  z  —  r  cos  <£. 

Hence  the  values  of  x,  y,  z  in  terms  of  r, 
(f>,  e  are 

x  =  r  sin  <£  cos  0,     y  =  r  sin  <f>  sin  0, 

z  =  r  cos  d>.  FlG-  2 


(1) 


Since  r  is  the  distance  from  0  to  P :  (x,  y,  z),  we  have  also  that 
(2)  ^  =  a?  +  3/2  +  «2. 

EXERCISES 

1.  Plot  the  points  (2,  90°,  180°),  (4,  60°,  -  30°),  (8,  f  TT,  f  TT). 

2.  Find  the  rectangular  coordinates  of  the  points  of  Ex.  1. 

3.  Find  the  spherical  coordinates  of   the  points   (0,  2,  0), 
(3,  4,  12),  (2,  —  2,  —  1),  checking  each  result  by  a  figure. 

*  It  is  to  be  noted,  however,  that  for  every  point  P  on  the  line  of  OA 
6  is  undetermined  and  that  for  0  in  particular  0  is  also  undetermined. 
Cf.  Ch.  X,  §  1. 


586  ANALYTIC   GEOMETRY 

4.  What  do  the  following  equations  represent  ? 

(a)  r  =  7;  (6)  <£  =  300;  (c)    0  =  f7r; 

(d)  tan  6  =  1 ;         (e)  tan  <£  =  1 ;         (/)  4  cos2  </>  =  1. 

5.  Find  the  equations  in  spherical  coordinates  of  the  follow- 
ing surfaces : 

(a)  The  sphere  of  radius  5,  center  at  0 ; 

(&)  The  meridian  half-plane  of  longitude  35° ; 

(c)  The  complete  plane  determined  by  this  half-plane ; 

(d)  The  upper  nappe  of  the  circular  cone  whose  vertex  is 
at  0,  whose  axis  is  along  OA,  and  whose  generating  angle  is 
60°; 

(e)  The  lower  nappe  of  the  cone  of  (d)  ; 
(/)  The  cone  of  (d). 

6.  What  do  the  following  pairs  of  equations  represent  ? 

(a)  r  =  3,  (£  =  120°;  (d)  r  =  3,  tan2<£  =  l; 

(6)  r  =  3,  0=f7r;  (e)   tan  0  =  2,  <f>  =  £*•; 

(c)  6  =  30°,  4  =  45° ;  (/)  tan  0  =  -  1,  cos2  </>  =f 

7.  Find  the  equations  in  spherical  coordinates  of  the  fol- 
lowing curves : 

(a)  The  small  circle  on  the  earth  of  colatitude  47° ; 
(6)  The  semicircle  on  the  earth  of  longitude  135° ; 

(c)  The  complete  circle  determined  by  the  semicircle  of  (6) ; 

(d)  The  ray  from  0  of  colatitude  60°  and  longitude  25°. 

8.  Determine  the  locus  of  each  of  the  following  equations : 
(a)  r  =  4  cos  <£ ;         (6)  r  —  6  sec  <£ ;         (c)  r  =  3  esc  <£. 

Find  the  equations  in  spherical  coordinates  of  the  following 
surfaces.     Identify  each  surface. 

9.  x*  +  f  +  z2  =  9.  13.   3x  +  2y  =  Q. 

10.  x>  +  /  -  fcV  =  0.  14.    3  z  -  4  =  0. 

11.  a2  +  f-  +  z2  =  4  y.  15. 

12.  4(x2  +  ?/2)  +  9  z2  =  36.  16. 


SPHERICAL  AND  CYLINDRICAL  COORDINATES     587 


FIG.  3 


2.  Cylindrical  Coordinates.  Given  a  point  0,  the  axis  of  z 
through  0,  and  the  plane  K  through  0  perpendicular  to  the 
axis  of  z.  In  K  introduce  a  system  of  polar  coordinates,  as 
shown.  Let  P  be  any  point  of  space  and  let  N  be  its  projec- 
tion on  K.  Then  the  polar  coordinates,  r  and  0,  of  N  and  the 
directed  line-segment  NP  =  z  deter- 
mine the  position  of  P.  The  three 
numbers,  taken  together,  are  known 
as  the  cylindrical  coordinates  (r,  0,  z) 
of  P. 

As  in  the  case  of  polar  coordinates 
in  the  plane  (Ch.  X,  §  1),  r  is  re- 
stricted to  be  positive  or  zero,  while 
0  is  unrestricted.  The  positive  direc- 
tion of  rotation  for  the  measurement 
of  6  is  as  indicated  in  the  figure. 

If  r0  (>  0),  #o,  ZQ  are  particular  values  of  r,  0,  z,  the  equation 
r  =  r0  represents  a  circular  cylinder  whose  axis  is  the  axis  of 
z ;  6  =  00  defines  a  half-plane  issuing  from  the  axis  of  z,  and 
z  =  z0  represents  a  plane  perpendicular  to  the  axis  of  z. 

Transformation  to  and  from  Rectangular  Coordinates. 
Choose  in  the  plane  K  the  Cartesian  axes  of  x  and  y  shown  in 
Fig.  3.  Referred  to  these  axes  and  the  axis  of  z,  P  has  the 
rectangular  coordinates  (x,  y,  z). 

It  is  clear  that  the  z  of  the  rectangular  coordinates  of  P  is 
precisely  the  z  of  the  cylindrical  coordinates  of  P.  The  for- 
mulas for  x,  y  in  terms  of  r,  6  and  for  r,  6  in  terms  of  x,  y 
are  those  of  transformation  in  a  plane  from  polar  to  rectangu- 
lar coordinates,  and  vice  versa  (Ch.  X,  §  6).  In  particular, 

(1)  x  =  r  cos  0,  y  =  r  sin  0, 

(2)  r2  =  a2  +  y2. 

EXERCISES 

1.  Plot  the  points  (2,  40°,  5),  (4,  -  f  TT,  -  3),  (0,  122°,  1). 

2.  Find  the  rectangular  coordinates  of  the  points  of  Ex.  1. 


588  ANALYTIC   GEOMETRY 

3.  Find  the  cylindrical  coordinates  of  the  points  (3,  4,  8), 
(12,  —  5,  —  3),  (0,  0,  —  6),  checking  each  result  by  a  figure. 

4.  What  do  the  following  equations  represent  ? 

(a)     r  =  5;         (6)     0  =  225°;         (c)     tan  0  =  1. 

5.  Find  the  equations  in  cylindrical  coordinates  of  the  fol- 
lowing surfaces : 

(a)  The  circular  cylinder  of  radius  7  whose  axis  is  the  axis 
of  2; 

(6)  The  half-plane  bounded  by  the  axis  of  z,  the  angle  from 
OA  to  it  being  60°  ; 

(c)  The  complete  plane  determined  by  this  half-plane. 

6.  What  do  the  following  pairs  of  equations  represent  ? 

(a)  r=3,  0  =  -£7r;         (c)  2z  =  5,  0  =  120°; 
(6)  r  =  5,  z  =  -  6;  (d)  3z  -  8  =  0,  tan0  =  2. 

7.  Find  the  equations  in   cylindrical   coordinates   of   the 
following  curves : 

(a)  An  arbitrary  line  parallel  to  the  z-axis ; 
(6)  The  circle  of  radius  3,  whose  center  is  on  the  2-axis  and 
whose  plane  is  parallel  to  K  and  5  units  below  it ; 

(c)  An  arbitrary  ray  perpendicular  to  the  axis  of  z  and  issuing 
from  a  point  on  it ; 

(d)  The  line  of  this  ray. 

8.  Determine  the  locus  of  each  of  the  following  equations : 
(a)     r2  +  z2  =  9;         (6)     r  =  4sin0;         (c)  r  sin  0  =  5. 

Find  the  equations  in  cylindrical  coordinates  of  the  follow- 
ing surfaces.  Identify  each  surface. 

9.  Ex.  9,  §  1.  10.   Ex.  10,  §  1.  11.    Ex.  11,  §  1. 
12.    Ex.  13,  §  1.             13.    Ex.  14,  §  1.  14.    Ex.  15,  §  1. 

15.    3  (x2-  +  y-}  —  2  z-  =  6.         16.    xy  -f  yz  +  zx  =  0. 

17.  Prove  that  a  plane  through  OA  (Fig.  1)  or  Oz  (Fig.  3) 
is  represented  by  the  same  equation  in  both  spherical  and 
cylindrical  coordinates. 


SPHERICAL  AND  CYLINDRICAL  COORDINATES     589 

3.  Triply  Orthogonal  Systems  of  Surfaces.  Consider  the 
three  sets,  or  families,  of  planes  which  are  parallel  to  the  co- 
ordinate planes  of  a  Cartesian  system.  These  families  of 
planes  evidently  have  the  following  properties :  (a)  Through 
each  point  of  space  there  passes  just  one  plane  of  each  family ; 
(6)  Two  planes  of  different  families  intersect  at  right  angles. 
We  say,  then,  that  the  three  families  of  planes  form  a  triply 
orthogonal  system  of  planes. 

The  equations  of  the  families  are,  respectively, 

(1)  x  =  k,  y  =  I,  z  =  m, 

where  k,  I,  m  are  arbitrary  constants,  or  parameters,  each  tak- 
ing on  any  value,  positive,  zero,  or  negative. 

Since  through  a  point  P:  (x0,  y0,  z0)  there  pass  just  three 
planes,  one  from  each  family,  namely  the  planes  x  =  x0,  y—  y0, 
z  =  z0,  and  since,  further,  P  is  the  only  point  which  the  three 
planes  have  in  common,  the  position  of  P  can  be  thought  of  as 
determined  by  the  three  planes.  From  this  point  of  view, 
then,  the  basis  of  the  rectangular  coordinate  system  is  seen  to 
be  the  triply  orthogonal  system  of  planes  (1). 

In  the  case  of  a  system  of  cylindrical  coordinates,  consider 
the  three  families  of  surfaces  : 

(2)  r  =  k,  6  =  1,  z=m, 

where,  of  the  parameters  k,  I,  m,  k  cannot  be  negative  (or  zero), 
I  may  be  restricted  to  the  range  of  values  :  0  <  I  <  2?r,  and  m 
is  unrestricted. 

The  first  family  of  surfaces  consists  of  the 
circular  cylinders  with  the  axis  of  z  as  axis ; 
the  second  family  is  made  up  of  the  half-planes 
issuing  from  the  axis  of  z ;  and  the  third,  of 
the  planes  perpendicular  to  the  axis  of  z.  It 
is  easily  seen  that  through  each  point  of  space, 
with  the  exception  of  those  on  the  z-axis,  there 
passes  just  one  surface  of  each  family,  and 
that  two  surfaces  of  different  families  intersect  FIG.  4 


590  ANALYTIC  GEOMETRY 

orthogonally.     We  say,  then,  that  the  three  families  form  a 
triply  orthogonal  system  of  surfaces. 

•A  point  P:  (r0,  00,  z0),  not  on  the  z-axis,*  is  the  single  point 
of  intersection  of  the  three  surfaces,  r  =  r0,  6  =  60,  z  =  z0  which 
pass  through  it.  Thus  the  basis  of  cylindrical  coordinates  is 
the  triply  orthogonal  system  of  surfaces  (2). 

EXERCISE 

Write  the  equations  of  the  three  families  of  surfaces  peculiar 
to  a  spherical  coordinate  system.  Describe  each  family  and 
draw  a  figure  showing  three  surfaces,  one  from  each  family, 
their  curves  of  intersection  and  their  common  point.  Prove 
that  the  three  families  constitute  a  triply  orthogonal  system 
of  surfaces  and  show  that  this  system  can  be  considered  as  the 
basis  of  spherical  coordinates. 

4.   Confocal  Quadrics.     Consider  the  quadrics 


a  —          o  —          c  —  A; 

/•v.2  ,/2  «2 

(2)  -2—  +  -^  ---  —  =1,  c2  <  I  <  b\ 

«2  _  i^  52  _  i       i_G2 

(3)  —  -  ----  ^—  --  —  =  1,          V  <  m  <  a\ 
a2  —  ra     m  —  b*      m  —  c2 

where  a,  b,  c  are  positive  constants  such  that  .a  >  6  >  c  and 
k,  I,  m  are  parameters  subject  to  the  given  restrictions. 

The  surfaces  (1)  are  all  ellipsoids  ;  the  surfaces  (2),  all 
hyperboloids  of  one  sheet  opening  out  along  the  axis  of  z  ;  and 
the  surfaces  (3),  all  hyperboloids  of  two  sheets  cutting  the 
axis  of  x. 

The  coordinate  planes  are  the  principal  planes  of  all  the 

*  To  remove  this  exception,  add  to  the  family  of  cylinders,  r  =  k  >  0, 
the  axis  of  z,  r  =  0.  Through  a  point  P  of  this  axis  passes,  then,  one 
surface  each  from  the  first  and  third  families,  and  every  surface  of  the 
second.  However,  all  these  surfaces  have  but  the  one  point  P  in  com- 
mon and  hence  can  be  considered  as  determining  the  position  of  P. 


SPHERICAL  AND  CYLINDRICAL  COORDINATES     591 


surfaces.  It  is  readily  shown  that  the  sections  by  the  (x,  y)- 
plane  of  all  the  surfaces  are  confocal  conies,  the  common  foci 
being  at  the  points  ( ±  Va2  —  62,  0,  0).  Similarly,  for  the  sec- 
tions by  the  (z,  cc)-plane  of  all  the  surfaces,  and  for  the  sections 
by  the  (y,  z)-plane  of  the  surfaces  (1)  and  (2),  —  the  (y,  z)-plane 
does  not  cut  the  surfaces  (3).  This  property  of  the  surfaces 
(1),  (2),  (3)  is  expressed  b*y  calling  them  confocal  quadrics. 

It  can  be  shown  that  through  each  point  of  space,  with  the 
exception  of  those  in  the  coordinate  planes,  there  passes  just 
one  surface  of  each  type  and  that  two  surfaces  of  different 
types  intersect  orthogonally  all  along  a  curve.*  Consequently, 
the  confocal  quadrics  form  a  triply  orthogonal  system  of  surfaces. 

This  triply  orthogonal  system  differs  in  one  respect  from 
those  studied  in  §  3,  in  that  the  three  surfaces,  one  of  each 
type,  which  pass  through  a 
point  P  situated  in  a  given 
octant  intersect  not  only  in  P 
but  also  in  one  point  of  each 
of  the  other  octants ;  this  is 
clear  since  all  three  surfaces 
are  symmetric  in  each  coordi- 
nate plane.  Consequently,  in 
the  so-called  ellipsoidal  coordi- 
nate system  based  on  the  con- 
focal  quadrics  there  are  eight  points  with  the  same  coordinates. 
This  ambiguity  can  be  avoided,  however,  by  considering  only 
a  restricted  region  of  space,  for  example,  the  first  octant. 

The  equations  (1),  (2),  (3)  can  be  written  as  the  single 
equation 


FIG.  5 


(4) 


_  \         &2  _ 


=  1, 


where  X  is  arbitrary  except  that  it  shall  not  take  on  the  values 
c2,  &2,  a2.  If  X  <  c2,  equation  (4)  defines  the  surfaces  (1),  etc. ; 
finally,  if  X  >  a2,  (4)  has  no  locus. 

*  Cf .  Osgood,  Differential  and  Integral  Calculus,   p.  326. 


592  ANALYTIC  GEOMETRY 

EXERCISE 

.Show  that  the  equation 


where  A  is  a  parameter  not  taking  on  the  values  a2  and  62, 
represents  three  families  of  paraboloids  defined  by  the  in- 
equalities A  <  b2,  W  <  A  <  a2,  a2  <  A.  Describe  each  family 
and  show  that  the  sections  of  all  the  surfaces  by  the  common 
principal  planes  —  the  (y;  z)-  and  (z,  »)-planes  —  are  conf  ocal 
parabolas  with  the  axis  of  z  as  axis.  The  surfaces  are  known 
as  confocal  paraboloids.  They  form  a  triply  orthogonal  system. 

TRANSFORMATION  OF  COORDINATES 

5.  Transformation  to  Parallel  Axes.  To  transform  from  a 
system  of  rectangular  axes  to  a  new  system  of  axes  having  the 
same  directions  as  the  old,  but  with  a  dif- 
ferent origin,  consider  a  point  P  whose 
coordinates  with  respect  to  the  two  systems 
are,  respectively,  (a;,  y,  z)  and  (x'}  y',  2/). 
Then 


x  —  cc 

Z    =  Z'  +  Z0, 

,2N  x'  =  x  -x0.         y'  =  y  -  2/o, 

z'  =  z  —  z0, 

where  (x0,  y0,  z0)  are  the  coordinates  of  the  new  origin,  O '? 
referred  to  the  old  axes  ;  of.  Ch.  XI,  §  1. 

Example.     What  surface  is  represented  by  the  equation 

Completing    successively  the  squares  of  the  terms  in  x,  y, 
and  z,  we  have 

2(aj  -  I)2  +  3(y  +  I)2  -  4(z-  £)2  =  12. 
On  setting 

x  ==^  *c  •—  i  j        y  ^—  y  ~T~  J-j        £  —  i?  ~~  ~s-j 


TRANSFORMATION  OF  COORDINATES 


593 


that  is,  on  transforming  to  parallel  axes  with  the  new  origin 
at  the  point  (1,  —  1,  ^),  the  equation  becomes 

2z'2  +  3t/'2-4z'2=12. 

Equation  (3)  is  thus  seen  to  represent  a  hyperboloid  of  one 
sheet  whose  center  is  at  (1,  —  1,  ^)  and  whose  axes  are  parallel 
to  the  coordinate  axes.  The  hyperboloid  opens  out  in  the 
direction  of  the  axis  of  z  and  the  semi-axes  of  the  minimum 
ellipse  have  the  lengths  V6  and  2. 

EXERCISES 

Determine  and  draw  roughly  the  surface  represented   by 
each  of  the  following  equations. 
1.   z2-4z-6z 
2. 
3. 

4.  a;2  -  3  y2  +  z2  -  8  x  +  12  y  +  6  z  +  13  =  0. 

5.  3y2-f  4z2-f-4a-6*/  +  16z  +  27  =  0. 

6.  2  x2  +  4  ?/2  -T-  3  z2  -  8  x  -  24  y  —  30  z  -  19  =  0. 

7.  2x>-3f  +  z*  +  8x  +  18y-16z-3  =  0. 

8.  z24-2?/2  +  6z2-2a;-2?/  +  18z  +  9  =  0. 

9.  2z2-5?/2  +  3z2  +  202/  +  6z-47  =  0. 
10.   y*  —  2xz-2x-6y  +  2z  +  11  =  0. 


6.  Rotation  of  the  Axes.  Through  the  origin  0  of  the 
(right-handed)  system  of  (x,  y,  z)-axes,  choose  arbitrarily 
mutually  perpendicular  directed  lines  to 
serve  as  the  axes  of  a  new  (right-handed) 
system  of  coordinates  (#',  y',  z').  Let  the 
direction  angles  of  the  axis  of  x',  referred 
to  the  old  system,  be  al9  /3j,  ylt  let  those 
of  the  axis  of  y'  be  «2,  fa,  y2>  and  those 
of  the  axis  of  z',  «3,  /33,  ys. 

Let  an   arbitrary   point   P  of    space  FIG.  7 


usual 
three 


594  ANALYTIC   GEOMETRY 

have  the  coordinates  (x,  y,  z)  and  (a/,  y',  z')  with  respect  to  the 
two  systems.  Join  0  to  P  by  the  two  broken  lines  OMNP 
and  OM'N'P,  where 

OM  =  x,  MN=  y,  NP=z-,  OM'  =  xf,  M'N'  =  y',  N'P  =  z'. 
Then 

(1)  Proj.  OM  +  Proj.  MN  +  lProj.  NP 

=  Proj.  OM'  +  Proj.  M  N'  +  Proj.  N'P, 

no  matter  on  what  directed  line  the  projections  are  taken. 
Choosing  the  positive  axes  of  x,  y,  and  z  in  turn  as  this  directed 
line,  we  have 

x  =  x'  cos  «!  +  y'  cos  «o  +  z'  cos  03, 

(2)  y  =  x'  cos  &  +  y'  cos  /32  +  2'  cos  f}3, 
Z  =  X  COS  yi  +  y'  COS  y2  +  z'  cos  7s- 

Here  cos  ax,  cos/?!,  cosyj  —  the  coefficients  of  a;'  —  are  the 
direction  cosines  of  the  axis  of  x1 ;  cos  «2,  cos  /32,  cos  y2,  those  of 
the  axis  of  y' ;  and  cos  03,  cos  /33,  cos  y3,  those  of  the  axis 
of  z'.  Let  us  denote  these  direction  cosines,  for  the  sake  of 
brevity,  by  Xx,  m,  Vi,  Xa,  ^  v2,  X3,  /*3,  v3,  respectively. 

Since  these  triples  of  numbers  are  direction  cosines  and, 
moreover,  direction  cosines  of  three  mutually  perpendicular 
lines,  we  have 

V  +  /"-I2  +  Vi2  =  1,  Xi\2  +  p.lf*V  +  ViV2  —  0, 

(3)  X22  +  tf  +  v22  =  1,  X2X3  +  f^n.3  +  v2v3  =  0, 

V  +  /*32  +  V32  =  1,  AgXi  +  /I3ft!  +  V3Vi  =  0. 

Since  the  three  directed  lines  form  a  right-handed  system,  it 
follows  by  Ex.  19  at  the  end  of  Ch.  XVIII  that  the  determi- 
nant of  their  direction  cosines  has  the  value  plus  one : 

(4)  |X/Mv|  =  l. 

Equations  (3)  and  (4)  express  completely  the  fact  that  the 
three  given  lines  through  0  which  serve  as  the  new  axes  are 
directed,  mutually  perpendicular  lines  forming  a  right-handed 
system. 


TRANSFORMATION  OF  COORDINATES  595 

The  direction  cosines  of  any  one  of  the  three  directed  lines 
can  be  expressed  simply  in  terms  of  those  of  the  other  two 
(Exs.  17,  18  at  the  end  of  Ch.  Xyill) : 

(")    A2  =  P-S^I  —  Mi^s?          /^2  ==  ''sAi  —  ^iA3,          v2  ^= 

A3  ==  /*iv2  —  PWi)          P"S  ==  ^iX2  —  ^Xij          v3  ==  A]ju,2  —  A<J/II. 

Since  a1}  a2,  03  are  the  angles  which  the  axes  of  x',  y',  z' 
make  with  the  axis  of  x,  they  are  the  direction  angles  of  the 
axis  of  x  with  respect  to  the  new  axes.  Similarly,  fa,  /32,  fa 
and  yx,  y2,  ys  are  respectively  the  direction  angles  of  the  axes 
of  y  and  z,  referred  to  the  new  system.  Consequently,  the 
equations  of  transformation  from  the  new  axes  to  the  old  are 

x'  =  x  cos  KI  +  y  cos  fa  -f-  z  cos  y1? 

(6)  y'  =  x  cos  05  +  y  cos  fa  +  z  cos  y2, 
z'  —  x  cos  «3  +  ?/  cos  fa  +  z  cos  y3. 

The  direction  cosines  of  the  old  axes  with  respect  to  the  new 
are,  in  our  notation,  X1?  A2,  A3,  ^  /u,2,  />t3,  vu  v2,  v3.  It  is  clear 
that  between  these  three  triples  there  exist  relations  similar 
to  the  relations  (3),  (4),  (5)  for  the  original  triples.* 

The  accompanying  diagram  gives  equations 
(2)  and    (6)  in  skeleton.     Reading   across  we 
obtain  (2)  and  reading  down  we  get  (6).     Also,  x 
the  rows  give  the  direction  cosines  of  the  old 
axes  with  respect  to  the  new,  and  the  columns, 

7     Z         V\        V2        V3 

those  of  the  new  axes  with  respect  to  the  old. 

Example  1.     Transform  the  equation  of  the  surface 

(7)  13  x2  +  13  f  +  10  z1  +  8  xy  -  4  yz  -  4  xz  -  36  =  0 

to  new  axes  through  0,  whose  direction  cosines  are  respectively 

_121          2_1_2          22_J_i 

*  Of  the  new  equations  only  those  of  the  form  (3)  are  different  from  the 
old.  The  new  equation  (4)  is  obtainable  from  the  old  by  interchanging 
rows  and  columns  in  the  determinant.  Similarly,  if  in  the  present  equa- 
tions (5)  the  columns  are  written  as  rows,  the  result  is  the  new  equations 
(6). 


596  ANALYTIC   GEOMETRY 

Here  *  =  i(-    *'  +  2  y'  +  2  «'), 

y  =  i(     2*'-    </'  +  2z'), 
z  =  £(     2rf  +  2y'-     z'). 

Substituting  these  values  for  a;,  y,  z  in  (7)  and  simplifying  the 
result,  we  obtain 

as*  +  y'*  +  2  z'2  =  4. 

This  equation  represents  an  ellipsoid  of  revolution  about 
the-z'-axis  as  axis.  Hence  (7)  represents  an  ellipsoid  of  revo- 
lution whose  axis  is  the  line  through  0  with  the  direction 
components  2,  2,  —  1. 

Example  2.     What  surface  is  represented  by  the  equation 
(8) 


We  make  the  transformation  to  parallel  axes 
(9)  x  =  x'  +  x0,       y  =  y'  +  y0,       z  = 

aiming  to  choose  the  new  origin  (x0,  yQ,  z0)  so  that  in  the  equa- 
tion resulting  from  (8)  the  linear  terms  in  x',  y',  z'  do  not 
appear.  Substituting  the  values  of  x,  y,  z  given  by  (9)  into 
(8),  collecting  terms,  and  then  setting  the  coefficients  of  x',  y', 
and  z'  equal  to  zero,  we  obtain  the  equations  : 

47/0-  2z0-    5  =  0, 


a*  +      yo  —  5  zo  +  10  =  0.  . 

These  equations  have  a  unique  solution,  namely,  x0  =  1, 
y0  =  -  1,  z0  =  2. 

If  (8)  is  transformed  to  parallel  axes  with  the  new  origin  at 
the  point  (1,  —  1,  2)  thus  determined,  it  becomes 

13  x'2  +  13  y'2  +  10  z'2  +  8  x'y'  -  4  y'z'  -  4  x'z'  -  36  =  0. 

But  this  is  the  same  equation  in  x',  y',  z'  as  (7)  is  in  x,  y,  z. 
Hence  (8)  represents  an  ellipsoid  of  revolution  whose  center  is 
at  the  point  (1,  —  1,  2)  and  whose  axis  of  revolution  has  the 
direction  components  2,  2,  —  1. 


TRANSFORMATION  OF  COORDINATES  597 

Example  3.  Consider  an  equation  in  which  only  one  of  the 
terms  in  xy,  yz,  zx  is  present,  for  example,  the  equation 

(10)  2x"--y'2-z'>-2yz-4:X  +  6y  +  2z  +  2  =  Q. 

The  term  in  yz  in  this  equation  can  be  removed  by  rotating 
the  y-  and  z-axes  about  the  axis  of  x  through  a  suitable  acute 
angle  0  ;  that  is,  by  application  of  the  transformation 

(11)  x  =  x',        y  =  y'  cos  6  —  z'  sin  0,        z  =  y'  sin  0  +  z'  cos  6. 

According  to  Ch.  XII,  §  5,  the  desired  angle  0  is  45°.  Trans- 
forming (10)  by  the  rotation  of  axes  (11),  where  6  =  45°,  we 
obtain 

x'2  _  y<2  _  2  x'  +  2  V2  y'  -  V2  z'  +  1  =  0. 

This  equation  can  be  written  in  the  form 

(X>  _  1)2  _  (y>  _  V2)2  =  V2(z'  -  V2), 
and  hence  becomes 


when  referred  to  axes  through  the  point  (1,  V2,  V2)  parallel 
to  the  axes  of  05',  y',  z'. 

It  follows,  then,  that  (10)  represents  a  hyperbolic  paraboloid 
whose  vertex,  referred  to  the  (xf,  y',  z')-axes,  is  at  the  point 
(1,  V2,  V2)  and  whose  axis  is  parallel  to  the  axis  of  z'. 

Remark.  The  general  method  of  procedure  to  determine 
the  surface  represented  by  an  equation  of  the  second  degree  in 
x,  y,  z  is  that  of  Examples  1  and  2  ;  the  equation  is  first  trans- 
formed by  a  change  of  origin  to  remove  the  linear  terms  in 
x,  y,  z  and  is  then  subjected  to  a  rotation  of  the  axes  to  get 
rid  of  the  terms  in  xy,  yz,  zx.  This  method  cannot  be  applied, 
however,  to  equation  (10)  of  Example  3,  for  it  is  impossible  to 
transform  (10)  so  that  the  linear  terms  disappear,  since,  if  this 
were  possible,  the  surface  would  be  symmetric  in  the  new 
origin  (Ex.  11  at  the  end  of  the  chapter),  whereas  we  know 
that  a  paraboloid  has  no  point  of  symmetry.  Accordingly. 
in  Example  3  and  in  similar  cases,  the  axes  are  first  rotated  to 


598  ANALYTIC   GEOMETRY 

remove  the  terms  in  xy,  yz,  zx  and  then  a  proper  change  of 
origin,  as  suggested  by  the  new  equation,  is  made. 

EXERCISES 

1.  Find  the  equations   of  the  rotation  of  the  axes  which 
introduces  the  two  directed  lines  through  0  with  the  direction 
cosines  ^-,  %,  —  f  and  -f-,  ^,  ^  as  the  axes  of  x'  and  y'  respectively. 

2.  Find  the  equations  of  a  rotation  of  the  axes  which  in- 
troduces the  planes, 


as  the  (a/,  y')-,  (y',  z')-,  and  (z,  #')-  planes,  respectively. 

3.    Find  the  equations  of  a  rotation  of  the  axes  which  in- 
troduces the  planes 


as  the  (x',  y')-  and  (y',  z')-  planes. 

4.  Transform  the  equation  of  the  hyperbolic  paraboloid 

a;2  —  y2  =  2  raz 

by  a  rotation  of  the  x-  and  y-axes  through  an  angle  of  —  45° 
about  the  axis  of  z.  Ans.   x'y'  =  mz'. 

Determine  the  surface  represented  by  each  of  the  following 
equations. 

5.  5 

6.  3o;2 

7.  x2- 


8.  Transform  the  equation  of  the  surface 

5z2  -  2y2  +  11  z2  +  12  xy  +  12yz  -  14  =  0 

by  the  rotation  of  the  axes  of  Ex.  1.     Thus  identify  the  sur- 
face. 

9.  The  equation 

13z2  -  ±xy  -  lOi/z  -  4zz  -  36y  +  36z  =  0 


TRANSFORMATION  OF  COORDINATES  599 

represents  a  central  quadric  whose  axes  have  the  direction 
components  2,  —  1,  —  1,  0,  1,  —  1,  1,  1,  1.  Identify  the 
quadric. 

10.    The  equation 


represents  a  paraboloid  whose  principal  planes  are  x  +  y  =  0, 
x  —  y  +  z  =  0.  Identify  the  paraboloid. 

11.  Show  that  the  equation  of  a  sphere  whose  center  is  at 
the  origin  is  not  changed  by  any  rotation  of  the  axes.  Actu- 
ally carry  through  the  transformation. 

7.  The  General  Equation  of  the  Second  Degree.    We  have 
defined  a  quadric  surface  as   any  surface  represented  by  an 
equation  of  the  second  degree  in  x,  y,  z,  that  is,  by  an  equation 
of  the  form  : 
(1)       Art  +  Rf  +  Cz"-  +  2  A'yz  +  2  B'xz  +  2  C'xy 

+  2A"x  +  2B"y  +  2  C"z  +  F=  0. 

We  propose  now  to  ascertain  whether  there  are  types  of 
quadric  surfaces  other  than  those  already  discussed  in  Chs. 
XXII,  XXIII,  and  to  sketch  a  method  whereby  the  type  of 
surface  defined  by  a  given  equation  of  the  form  (1)  can  be 
determined. 

As  in  the  corresponding  problem  in  the  plane  (Ch.  XII), 
transformations  of  coordinates  play  an  important  role.  In 
particular,  the  expressions  formed  from  the  coefficients  of  (1), 
which  are  invariant  (Ch.  XII,  §  6)  under  any  change  of  axes, 
are  fundamental.  Chief  among  these  invariant  expressions 
are  the  determinants 

A      €'     B1     A" 


D  = 


C"       Pi        A'       R" 

r»    7i     A>  \  - 

0       B      A   ,  A-       ,  ,  „„ 

B'    ^1'     (7 


A"    B"     C"    F 

which  correspond  to  the  invariants  B2  —  4  AC  and  A  in  the 
case  of  the  general  equation  of  the  second  degree  in  x  and  y 
(Ch.  XII,  §  6). 


600  ANALYTIC   GEOMETRY 

We  state,  without  proof,  the  following  theorems : 

THEOREM  1.  If  equation  (1)  represents  a  surface,  and  if 
D  =£  0,  the  surface  is  symmetric  in  just  one  point.  If  D  =  0, 
there  is  in  general  no  point  of  symmetry,  and  when  there  is  one, 
there  are  infinitely  many. 

THEOREM  2.  If  equation  (1)  represents  a  surface  and  if 
D  3=  0,  the  surface  is  symmetric  in  three  mutually  perpendicular 
planes,  and  these  are,  in  general,  all  the  planes  of  symmetry.  If 
D  =  0,  it  is  symmetric  in  two  perpendicular  planes  and  these 
are,  in  general,  all  the  planes  of  symmetry. 

It  is  clear  from  these  theorems  that,  in  discussing  equation 
(1),  two  essentially  different  cases  arise,  according  as  D  =£  0 
or  D  =  0.  - 

Case  1.  D  =£  0.  A  surface  defined  by  an  equation  of  the 
form  (1)  for  which  D  =£  0  is  symmetric  in  a  unique  point  0', 
by  Th.  1.  The  coordinates  of  0'  can  be  found  by  the  method 
of  Example  2,  §  6.  A  transformation  to  parallel  axes  with 
the  new  origin  at  0'  removes  the  linear  terms  in  (1),  leaves  the 
quadratic  terms  unchanged,  and,  as  can  be  shown,  makes  the 
constant  term  into  A/Z>.  Thus  (1)  becomes 

(2)   Ax'*  +  By'2  +  Cz'2  +  2  A'y'z'  +  2  B'x'z'  +  2  C'x'y'  +  A/Z)=0. 

Since  D  3=  0,  the  surface  is,  by  Th.  2,  symmetric  in  three 
mutually  perpendicular  planes  whose  common  point,  since  it 
is  a  point  of  symmetry,  must  be  0'.  To  determine  from  equa- 
tions (1)  or  (2)  the  precise  positions  of  these  planes  through 
0'  is  a  problem  of  intrinsic  difficulty  which  we  shall  not 
attempt  to  discuss.  When  once  the  positions  are  known,  how- 
ever, a  rotation  of  the  axes  which  brings  the  coordinate  planes 
into  coincidence  with  them  serves,  either  immediately  or 
eventually,  to  remove  the  terms  in  y'z',  z'x',  and  x'y'  in  (2); 
cf.  Ex.  12  at  the  end  of  the  chapter.  We  obtain,  then,  the 
final  equation 

(I)  ax'"2  +  by'"-  +  cz"2  +  A/D  =  0, 


TRANSFORMATION  OF  COORDINATES  601 

For  this  equation,  D  =  abc  and  hence,  since  D  =£  0,  no  one  of 
the  coefficients  a,  b,  c  can  be  zero. 

If  A  =£  0,  (I)  and  hence  (1)  represents  a  central  quadric  (an 
ellipsoid  or  hyperboloid)  or,  in  case  a,  b,  c,  A/Z)  are  all  of  the 
same  sign,  it  has  no  locus.* 

If  A  =  0,  (I)  and  hence  (1)  represents  a  cone,  or,  in  case  a,  b,  c 
are  all  of  the  same  sign,  a  point. 

Case  2.  D  =  0.  A  surface  defined  by  an  equation  of  the 
form  (1)  for  which  D  =  0  has  in  general  no  point  of  symmetry 
(Th.  1),  and  hence  it  is  in  general  impossible  to  transform  to 
parallel  axes  so  that  the  linear  terms  in  (1)  disappear.  There 
are,  however,  at  least  two  mutually  perpendicular  planes  of 
symmetry,  by  Th.  2.  If  the  positions  of  two  such  planes  are 
known,  a  rotation  of  the  axes  whereby  two  of  the  coordinate 
planes  become  respectively  parallel  to  them  serves,  either  im- 
mediately or  eventually,  to  remove  the  terms  in  yz,  zx,  and  xy 
in  (1) ;  cf.  Ex.  13  at  the  end  of  the  chapter.  Thus  (1)  becomes 

(3)       ax'2  +  by'2  +  cz'2  +  2  a"x'  +  2  b"y'  +  2  c"z'  +  F=Q. 

For  this  equation,  D  =  abc,  and,  since  D  —  0,  abc  =  0.  Now 
a,  b,  c  are  not  all  zero,  since  otherwise  (3),  and  hence  (1), 
would  not  be  a  quadratic  equation.  Two  cases  then  arise, 
according  as  one  or  two  of  the  coefficients  a,  b,  c  vanish. 

A.  One  of  the  coefficients  a,  b,  c  vanishes.  Since  (3)  bears 
equally  on  x',  y',  z'  it  is  immaterial  which  one  of  the  coefficients 
a,  b,  c  we  assume  to  be  zero.  Suppose  that  c  =  0 : 

ax'2  +  by'2  +  2  a"x'  +  2  b"y'  +  2  c"zr  +  F=Q,    ab^O. 

By  a  change  of  origin  to  the  point  (—  a" /a,  —  b"/b,  0),  this 
equation  becomes 

(Ha)  ax"2  +  by"2  +  2  c'Y  +  /=  0,  ab  =£  0. 

*  In  reducing  (1)  to  the  form  (I),  it  was  assumed  that  (1)  represents 
a  surface  ;  the  method  of  reduction  is  quite  the  same,  however,  if  (1)  has 
no  locus.  Similarly,  the  method  of  reduction  in  case  2  is  always  applica- 
ble, both  when  (1)  represents  a  surface,  as  assumed,  and  when  it  does  not. 


602  ANALYTIC   GEOMETRY 

For  (Ila),  A  =  —  abc"2  and  hence  A  =£  0  or  A  =  0,  according 
as  c"  ^=  0  or  c"  =  0.  If  c"  =£  0,  a  change  of  origin  to  the  point 
(0,  0,  -//2c")  reduces  (Ila)  to 

ax"2  +  by"2  +  2  c"z"  =  0,  a&  =£  0. 

If  c"  =  0,  (Ha)  becomes 

ax"2  +  by"2  +/=  0,  ab  =£  0. 

Hence  we  conclude  the  following : 

Jjf  A  =£  0,  (Ha)  and  Aewce  (1)  represents  a  paraboloid  (elliptic 
or  hyperbolic). 

If  A  =  0,  (Ha)  and  hence  (1)  represents,  in  the  case  /=£  0, 
an  elliptic  or  hyperbolic  cylinder,  or  it  has  no  locus;  if  f=  0,  it 
represents  two  intersecting  planes  or  a  line. 

B.  Two  of  the  coefficients  a,  b,  c  vanish.  Here  again  it  is 
immaterial  which  two  of  the  three  coefficients  we  assume  to 
be  zero.  Suppose  that  b  =  c  =  0 : 

ax'2  +  2 a"x'  +  2b'y'  +  2 c"z'  +  F  =  0,  a  =£  0. 

By  a  change  of  origin  to  the  point  (—  a"/a,  0,  0)  and  by  a 

proper  rotation  of  the  axes  about  the  axis  of  x',  this  equation 

becomes 

(116)  ax"2  +  2  dz"  +  /=  0,  a  =£  0. 

Here  A  is  always  zero.  Equation  (116),  and  hence  (1),  repre- 
sents a  parabolic  cylinder,  if  d  =£  0;  if  d  =  0,  it  represents  two 
parallel  planes,  a  single  plane,  or  has  no  locus. 

Summary.  The  new  types  of  loci  of  equations  of  the  form 
(1)  which  have  resulted  from  this  investigation,  are: 

i)  A  point,  —  which  is  a  limiting  form  of  an  ellipsoid  and  is 
frequently  spoken  of,  in  this  connection,  as  a  null  ellipsoid.  It 
is  to  be  noted,  from  the  discussion  of  (I),  that  the  correspond- 
ing limiting  form  of  a  hyperboloid  is  a  cone. 

ii)  Two  intersecting  planes,  a  line,  two  parallel  planes  or  a 
single  plane,  all  of  which  are  limiting  forms  of  cylinders.  We 
shall  call  them  degenerate  cylinders;  two  intersecting  planes, 
a  degenerate  hyperbolic  cylinder ;  a  line,  a  degenerate  (or  null) 


TRANSFORMATION  OF  COORDINATES 


603 


elliptic  cylinder ;  two  parallel  planes,  or  a  single  plane,  a  de- 
generate parabolic  cylinder. 

We  can  now  summarize  our  results  : 

THEOREM  3.  An  equation  of  the  form  (1),  if  it  has  a  locus, 
represents  a  central  quadric,  a  paraboloid,  a  cone  or  a  point,  or 
a  cylinder  (non-degenerate  or  degenerate). 

The  following  table  shows  when  each  of  the  four  cases 
occurs  and  thus  furnishes  a  means  of  determining  the  type  of 
surface  denned  by  any  given  equation  of  the  form  (1). 


A=jfc  0 

A  =  0 

D^fcO 

Central  quadric 
or  no  locus 

Cone 
or  point 

D  =  0 

Paraboloid 

Cylinder  or 
no  locus 

Equations  (Ila)   and  (116)    can   be   written   as   the   single 
equation 
(II)  oaf-  +  by*-  +  2  dz  +f=  0,  a  *  0, 

where  the  primes  have  been  dropped  from  the  variables.     We 
have  then,  in  conclusion,  the  theorem : 

THEOREM  4.     An  equation  of  the  form  (1)  can  be  reduced  by 
transformations  of  coordinates  to  one  of  the  two  forms : 

(I)  ace2  +  by"2  +  cz2  +  /  =  0,  abc  =£  0, 

(II)  ace2  +  by12  +  2  dz  +/=  0,  a  =£  0. 

EXERCISES 

Determine  to  which  of  the  four  types  the  quadric  surface 
represented  by  each  of  the  following  equations  belongs. 
l.   That  of  Ex.  7,  §  6.  2.   That  of  Ex.  8,  §  6. 

3.  31  x>-  +  41  y*  -  23  z>-  +  48  yz  +  72  xz  -  24  xy 

-  72  ce  -  48  y  +  46  z  -  23  =  0. 

4.  2  x*  +  2  f-  -(-  2  z1  +  yz  +  xz  +  xy  —  4  ce  —  y  —  z  -  4  =  0. 


604  ANALYTIC   GEOMETRY 

5. 


-  26x  -  Uy  -  18z  -  18  =  0. 
6.    7  x2  -f  7  y1  +  4  z"-  —  8  yz  +  8  zz  —  2  a*/ 


7.  The  surface   (3)  is  symmetric  in  three  planes  whose 
normals   have   the   direction    components   6,  —  3,  2,     2,  6,  3, 
—  3,   —  2,  6.     Determine  the  precise  nature  and  position  of 
the  surface. 

8.  The  surface  (4)  is  symmetric  in  three  lines  whose  direc- 
tion components  are  2,  —  1,  —  1,     0,  1,  —  1,     1,  1,  1.     De- 
termine its  precise  nature  and  .position. 

9.  The   surface    (5)    is   symmetric   in   two   planes   whose 
normals  have  the  direction  components  1,  —  1,  —  2,     3,  1,  1. 
Determine  its  precise  nature  and  position. 

10.    Two  principal  planes  of  the  surface   (6)  are   parallel 
respectively  to  the  planes 

x  —  y  +  z  =  0,         a?  +  y  =  0. 
Determine  the  precise  nature  and  position  of  the  surface. 

EXERCISES  ON   CHAPTER  XXIV 

1.  Prove  that  a  curved  surface  whose  equation  in  spherical 
coordinates  does  not  contain  r  is  a  cone  with  the  pole  as  vertex. 

2.  Show  that  a  curved  surface  whose  equation  in  spherical 
coordinates  does   not   contain   6  is   a   surface  of   revolution. 
What  is  its  axis  ? 

3.  Prove  that  a  curved  surface  whose  equation  in  cylindri- 
cal coordinates  does  not  contain  z  is  a  cylinder. 

4.  Show  that  a  curved  surface  whose  equation  in  cylindri- 
cal coordinates  does  not  contain  0  is  a  surface  of  revolution. 

TRANSFORMATION  OF  AXES 

5.  Prove  that  the  transformation  to  new  axes  through  0 
whose  direction  cosines  are  —  -|,  f  ,  f  ,     •§,  —  £,  •§•  ,     f  ,  f  ,  —  ^  is 


TRANSFORMATION  OF  COORDINATES  605 

identical  with  a  rotation  of  the  original  axes  about  the  line 
x  =  y  =  z  through  180°. 

Suggestion.  Show  that  the  equations  of  transformation  are 
equivalent  to  those  connecting  the  coordinates  of  two  points 
symmetric  in  the  line. 

6.  Find  the  equations  of  the  •  transformation  which  intro- 
duces as  axes  the  three  mutually  perpendicular  lines  through 
the  point   (x0,  y0,   z0)    with  the   direction  angles    al}   fa,   y1} 

«2>  fa,  72>       «3>  fa)   73- 

7.  Set  up  the  equations  of  a  rotation  of  the  axes  which 
introduces  the  plane  x  +  y  +  z  =  Q&s  the  (a;,  y)-plane. 

8.  Determine  the  precise  nature  of  the  curve  of  intersec- 
tion of  the  plane  x  +  y  +  z  =  0  with  the  surface 

x2  —  xy  +  yz  —  zx  —  x  —  y  —  2  =  0. 


Suggestion.     Use  the  result  of  Ex.  7. 
9.    A  line  of  symmetry  of  the  surface 
xy  +  yz  +  xz  =  2 

is  the  line  x  =  y  =  z.     Determine  the  precise  nature  of  the 
surface. 

10.  A  plane  of  symmetry  of  the  surface 

xz  +y2  +  z2  +  xy  +  yz  —  xz  —  x  +  y  —  z  =  0 

is  the  plane  x  —  y  —  2  z  =  0.     What  is  the  exact  nature  of  the 
surface  ? 

11.  Show  that,  if  a  quadric  surface  is  symmetric  in  the  ori- 
gin, its  equation  contains  no  linear  term  in  x,  y,  z,  and  con- 
versely. 

12.  A  quadric  surface  is  symmetric  in  each  of  the  coordinate 
planes.     Prove   that   either   the   equation   of   the   surface   is 
of  the  form 

ax*  +  by*  +  cz-  =  d, 

or  that  the  surface  consists  of  two  coordinate  planes.     Show 
that  in  the  latter  case  the  equation  can  be  reduced  to  the 


606  ANALYTIC   GEOMETRY 

desired  form  by  rotating   two  of  the   axes   about   the   third 
through  an  angle  of  45°. 

13.  A  quadric  surface  is  symmetric  in  two   planes  which 
are  parallel  to  or  identical  with  two  coordinate  planes.     Show 
that  either  the  terms  in  yz,  zx,  xy  do  not  appear  in  the  equation 
of  the  surface  or  the  surface  itself  consists  of  two  planes  of 
the  type  described.     Prove  that  in  the  latter  case  the  terms 
in  yz,  zx,  xy  can  be  removed  from  the  equation  by  rotating  two 
of  the  axes  about  the  third  through  an  angle  of  45°. 

14.  Show  that,  if  a  quadric  surface  is  symmetric  in  a  coordi- 
nate plane  or  in  a  plane  parallel  to  a  coordinate  plane,  its 
equation  contains,  in  general,  at  most  one  of  the  three  terms 
in  yz,  zx,  xy.     When  does  the  exception  occur  ? 

15.  Prove  that  the  conclusion  of  the  previous  exercise  fol- 
lows if  the  surface  is  symmetric  in  a  coordinate  axis  or  in  a 
line  parallel  to  a  coordinate  axis.     When  does  the  exception 
occur  in  this  case  ? 


INDEX 


Abridged  notation,  171 ; 

problems  in  — ,  189,  287. 
Absolute  value,  4 ; 

use  of  — ,  11,  41,  42,  5lf  466. 
Acoustical    property    of     conies,    cf. 

Focal  property. 
Affine  transformations,  342 ; 

homogeneous  — ,  348,  355  ; 

singular  — ,  358. 
Anchor  ring,  544. 

Angle,   in  the  plane,   positive  sense 
for  measurement  of,  13 ; 

slope  — ,  13 ; 

—  between  two  lines,  38. 

Angle,  in  space,   between  two  lines, 

407,  425,  435,  442 ; 
direction  —  of  a  directed  line,  420 ; 

—  between  two  planes,  455 ; 

—  between  a  line  and  a  plane,  487. 
Asymptotes,  129,  259. 
Asymptotic  cone,  543,  552,  557. 
Asymptotic   planes  of   a   hyperbolic 

cylinder,  533. 

Axes  of  a  conic,  88,  102,  125,  127 ; 
equations  of  — ,  260 ; 
construction  of  — ,  324. 
Axes  of  a  quadric  surface,  533,  549, 

551,  555. 

Axes  of  coordinates,  7,  410. 
Auxiliary  circle  of  an  ellipse,  120 ; 

—  of  a  hyperbola,  143. 

Bisectors    of    angles,    between    two 
lines,  281,  287; 

—  between  two  planes,  522. 
Brennpunkt,  97. 

Cardioid,  201,  209,  215,  286. 

Cassini,  ovals  of,  215. 

Center    of    a    conic,    102,    125,    247, 

259; 

construction  of  — ,  324. 
Center    of    a   quadric    surface,    543, 

549,  551. 


Central  conies,  299,  313. 
Central  line  of  a  strain,  304. 
Central  quadrics,  575,  601. 
Circle,  see  Contents,  Ch.  IV ; 

null  — ,  67  ; 

equation    of    —    in    determinant 
form,  394,  398 ; 

nine-point  — ,  78 ; 

—  as  limit  of  ellipses,  103,  140 ; 
parametric    representation    of   — , 

119; 

common  chords  of  three  — ,  170 ; 

equations  of  —  in   polar   coordi- 
nates, 194,  211; 

—  inscribed  in  a  triangle,  283 ; 
four  points  on  a  — ,  395,  398. 

Circle  in  space,  529 ; 
tangent  line  to  — ,  545 ; 
as   section   of    a   quadric   surface, 

564. 

Cissoid,  214. 

Coefficient  of  a  strain,  309. 
Colatitude,  584. 
Columns  and  rows  of  a  determinant, 

363; 

interchanges  of  — ,  376,  380. 
Compatibility  of  n-\-l   equations  in 
n    unknowns,     384,     401,     403, 
510. 
Compression,      simple,      cf.     Strain, 

one-dimensional. 
Conchoid,  215. 
Condition  that 

the  roots  of  a  quadratic  equation 

be  equal,  175 ; 

two    sets   of   numbers    be    propor- 
tional, 385; 

an  equation  be  homogeneous,  538. 
Condition,  in  the  plane,  that 
a  point  lie  on  a  curve,  19  ; 
two   points   be   collinear  with   the 

origin,  25,  402 ; 

three     points     be     collinear,     54, 
392 ;   problems,  60-63,  394 . 


607 


608 


INDEX 


four  points  lie  on  a  circle,  395,  398 ; 
six  points  lie  on  a  conic,  398 ; 
two  lines  be  parallel,  36,  45,  46, 

402; 
two    lines    be     perpendicular,    37, 

45;  46; 

two  lines  be  identical,  46,  402 ; 
three  lines  be  concurrent,  53,  189, 

393;    problems,  54-60,  62,  189, 

394; 
a  line  be  orthogonal  to   a  circle, 

76; 

two  circles  be  orthogonal,  77; 
a  line  be  tangent  to  a  conic,  183, 

192,  403 ; 
two  diameters  be  conjugate,  291, 

296; 

a  conic  be  degenerate,  257,  403 ; 
a  transformation  be  isogonal,  348. 
Condition,  in  space,  that 

a  point  lie  on  a  surface,  445 ; 

a  point  lie  on  a  curve,  471 ; 

two  points  be  collinear  with  the 

origin,  441 ; 

three  points  be  collinear,  505 ; 
four  points  be  coplanar,  508 ; 
five  points  lie  on  a  sphere,  527 ; 
two   lines   be   perpendicular,   426, 

435; 
two    lines    be    parallel,    420,    427, 

430,  432,  436 ; 
two  lines  intersect,  512 ; 
three  lines  be  parallel  to  a  plane, 

440; 
the    normals    to    four    planes    be 

parallel  to  a  plane,  508; 
three  lines  be  mutually  perpendic- 
ular, 594 ; 

a  line  lie  in  a  plane,  506 ; 
a  line   be   tangent  to   a   quadric, 

567; 

two  planes  be  perpendicular,  par- 
allel, identical,  455 ; 
three  planes  pass  through  a  line, 

504; 
four  planes  pass  through  a  point, 

510; 
a  plane  be  orthogonal  to  a  sphere, 

545; 
a  line  be  orthogonal  to  a  sphere, 

546: 
two  spheres  be  orthogonal,  546 ; 


a  diameter  and  a  diametral  plane 

be  conjugate,  571 ; 
three     diameters     be     conjugate, 

573; 

three  diametral  planes  be  conju- 
gate, 573 ; 
three  numbers  be  direction  cosines, 

422; 

a     directed     trihedral     be     right- 
handed,  443. 
Cones,  536. 

Confocal  conies,  145,  148. 
Confocal  parabolas,  95,  146. 
Confocal  paraboloids,  592. 
Confocal  quadrics,  590. 
Conies    as    sections,    of    a    circular 
cone,  144 ; 

—  of  a  quadric  cylinder,  534 ; 

—  of  a  quadric  surface,  562. 
Conies,  definition  of,  144 ; 

equations  of  —  in  polar  coordi- 
nates, 202,  210,  211; 

—  as    loci    of    equations    of    the 
second  degree,  257 ; 

equation  of  —  through  five  points, 

395,  398 ; 

six  points  on  a  — ,  398,  399. 
Conies,     degenerate,     definition     of, 

257; 
equations    of   — ,    191,    244,    253, 

260,  402 ; 
examples  of  — ,  237,  246,  254,  259 ; 

—  through  five  points,  396,  397 ; 
condition  for  — ,  257,  402 ; 

—  as   intersections   with   quadric 
surfaces,  256,  257,  563,  569. 

Conies,  similar  and  similarly  placed, 
260; 

—  —  as  sections  of  two  quadric 
surfaces,  535,  563,  580 ; 

as  parallel  sections  of  a  quad- 
ric surface,  550,  553,  555,  562- 
564,  580. 

Conjugate  diameters  and  diametral 
planes,  571-574,  579. 

Conjugate  diameters  of  a  conic, 
see  Contents,  Ch.  XIV. 

Conjugate  diameters  of  a  quadric 
surface,  572-574,  579. 

Conjugate  diametral  planes,  572- 
574,  579. 

Conjugate  hyperbolas,  141,  298. 


INDEX 


609 


Conjugate  hyperbolic  cylinders,  533. 
Conjugate  hyperboloids,  551 ; 

—  of  revolution,  542 ; 
sections  of  — ,  563. 

Construction,'  geometrical,   of  a   pa- 
rabola, 88; 

—  of  an  ellipse,  104,  114,  119; 

—  of  a  hyperbola,  124,  138,  143 ; 

—  of  center,   axes,  foci,   tangents 
of  a  conic,  324,  325 ; 

—  of  poles  and  polars,  319,  320, 
325.. 

Constructions,     mechanical,      of     a 
parabola,  147 ; 

—  of  an  ellipse,  101,  147 ; 

—  of  a  hyperbola,  147. 
Continued  equality,  cf.  Equality. 
Coordinate  planes,  410. 
Coordinates,    rectangular,  7,  14 ;    in 

space,  409,  411,  589; 
polar — ,193; 

spherical  — ,  584,  590,  604 ; 
cylindrical  — ,  587,  589,  604 ; 
ellipsoidal  — ,  591. 
Cramer's  rule,  381,  464. 
Cubic     equations,     graphs     of,     85, 

331,  353,  354. 
Cubic,  twisted,  492,  494. 
Curves,  in  the  plane,  definition  of, 

19; 
plotting   of — ,    19,  86,   331,    353; 

in  polar  coordinates,  198 ; 
symmetry  of  — ,  84-87 ;    in  polar 

coordinates,  201. 

Curves,  in  space,  definition  'of,  471 ; 
parametric    representation    of    — , 

490; 
Cylinders,  equations  of  special,  444, 

446; 

definition   and    equations   of   gen- 
eral — ,  532. 
Cylindrical  coordinates,  587, 589,  604. 

Deformation  of  an  elastic  body,  357. 

Degenerate  conies,  cf.  Conies. 

Descartes,  1. 

Determinant    of    a    transformation, 
358. 

Determinants,     see     Contents,     Ch. 

XVI; 

applications  of  —  to  linear  equa- 
tions, 381-391,  401,  403; 


applications  of  —  to  plane  ana- 
lytic geometry,  391-399,  401- 
403; 

applications  of  —  to  solid  ana- 
lytic geometry,  438,  440,  443, 
453,  463,  467,  477,  489,  496, 
504,  507-511,  512-515,  517, 
518,  526,  594,  599. 

Diagonals  of  a  determinant,  363. 

Diameters  of  a  conic,  see  Contents, 
Ch.  XIV. 

Diameters  of  a  quadric  surface, 
569-574,  579. 

Diametral  planes  of  a  quadric  sur- 
face, 570-574,  579. 

Diocles,  cissoid  of,  214. 

Directed  line-segments,  cf.  Line- 
segments. 

Directed  trihedrals,  430-432 ; 

right-handed  and  left-handed  — , 
443. 

Direction  angles  of  a  directed  line, 
420. 

Direction  components,  see  Contents, 
Ch.  XVIII. 

Direction  cosines,  see  Contents,  Ch. 
XVIII ; 

—  of   three   mutually   perpendic- 
ular lines,  594. 

Directrices  of  conies,  88,  116,  137. 
Directrix  of  a  cylinder,  532. 
Discriminant,   of  a  quadratic  equa- 
tion in  x,  175,  402 ; 

—  of  general  equation  of  second 
degree  in  x,  y,  248,  402,  403. 

Distance,  between  two  points,  10; 
in  polar  coordinates,  210;  in 
space,  414 ; 

—  of  a  point  from  a  line,  41 ;    in 
space,  514; 

—  between  two  parallel  lines,  51 ; 

—  of  a  point  from  a  plane,  460 ; 

—  between    two    parallel    planes, 
466; 

—  between    two    lines    in    space, 
515. 

Division  of  a  line-segment,  17,  416 ; 
harmonic  — ,  309,  359,  575. 

Eccentric  angle,  of  an  ellipse,  120, 
306; 

—  of  a  hyperbola,  143. 


610 


INDEX 


Eccentricity   of   a   conic,    102,    116, 

128,  137,  140. 

Element  of  a  determinant,  363. 
Ellipse,  see  Contents,  Ch.  VII ; 

equations    of    —    not    in    normal 

form,  114,  140; 

equations  /of  —  in   polar   coordi- 
nates, 203,  211 ; 

—  as  locus  of  equation  of  second 
degree,  239,  241,  246,  249; 

null  or  degenerate  — ,  245,  256 ; 
area  of  — ,  341. 
Ellipsoid,  548 ; 

sections    of    — ,    548,     550,     563, 

564-566,  569 ; 
volume  of  — ,  578 ; 
similar  — ,  581 ; 
null  — ,  602. 

Ellipsoids  of  revolution,  541,  548. 
Elongation,  simple,    cf.    Strain,  one- 
dimensional. 
Equality,   continued,   representing  a 

line  in  space,  473,  479,  482 ; 
reduction  of  —  to  normal  form,  480 ; 

—  equivalent    to    representation 
by  projecting  planes,  502. 

Equation,  general,  of  the  second 
degree  in  x,  y,  see  Contents, 
Ch.  XII. 

Equation,  general,  of  the  second  de- 
gree in  x,  y,  z,  599 ; 
reduction  of  —  in  special   cases, 

592,  595-597 ; 

reduction   of  — -  in   general   case, 
597,  600-603,  605. 

Equation  of  a  curve,  20. 

Equation  of  a  surface,  445. 

Equations,  linear,  cf.  Linear  equa- 
tions. 

Equations,  simultaneous,  of  a  space 
curve,  472. 

Equation  u+kv  =  0,  165,  188-190; 
in  space,  520. 

Equation  uv  =  0,  173,  190;  in  space, 
521. 

Equiangular  hyperbola,  cf.  Rec- 
tangular hyperbola. 

Equiangular  transformations,  cf. 
Isogonal  transformations. 

Factorization  of  a  transformation, 
343,  355 ; 


—  of   particular   transformations, 
349; 

exercises  in  — ,  342,  351,  357. 
Figures  in  plane  representing  space 

relations,  411. 
Focal  chords,  95,  98,  212. 
Focal   property   of   conies,    96,    108, 

134,  324. 

Focal  radii,  88,  106,  128,  212. 
Focus,  88,  101,  124 ; 

origin  of  name  — ,  97 ; 

construction  of  — ,  324. 

Generator,  cf.  Ruling. 

Graphs  of  equations,  in  rectangular 

coordinates,    18-20,   21,   83,   86, 

87,  331,  353,  354 ; 

—  in  polar  coordinates,  198. 

Harmonic  division,  309,  575 ; 

—  unchanged  by  an  affine  trans- 
formation, 359. 

Helix,  491. 

Homogeneity,  348,  537,  538. 

Homogeneous  linear  equations,  387, 

401. 

Hyperbola,  see  Contents,  Ch.  VIII ; 
equations    of    —    not    in    normal 

form,  140; 

equations  of  —  in  polar  coordi- 
nates, 203,  211 ; 

• —  as  locus  of  equations  of  the 
second  degree,  239,  241,  245, 
248,  '250 ; 

—  with    foci    on    the    axis    of   y, 
141,  221. 

Hyperboloid,    biparted    or    of    two 

sheets,  551 ; 
sections  of  — ,  551,  553,  561-563, 

565; 

similar  —  -.  581. 
Hyperboloid,    unparted    or    of    one 

sheet,  550 ; 
sections    of    — ,    550,    553,    561- 

563,  565,  566; 

asymptotic  cone  of  — ,  551,  557 ; 
rulings  of  — ,  555-559,  568,  581- 

583; 
parametric    representation    of   — , 

559; 
similar  — ,  581. 


INDEX 


611 


Hyperboloids     of     revolution,     541, 
550,  551. 

Identical  transformation,  342. 
Inequalities,  loci  of,  277,  279,  522. 
Infinity,  13. 
Initial  ray,  193. 
Intercepts,  of  a  line,  33 ; 

—  of  a  plane,  450. 
Intersection,  of  two  lines,  22 ; 

—  of  two  curves,  23 ; 

—  of  three  planes,  463 ; 

—  of  two  planes,  470,  476 ; 

—  of  a  line  and  a  plane,  488,  492 ; 
.     —  of  a  plane  and  a  sphere,  529 ; 

—  of  two  spheres,  530 ; 

—  of  a  curve  and  a  surface,  488, 
492; 

—  of  two  surfaces,  472 ; 

—  of  three  surfaces,  463. 
Invariants    of    general    equation    of 

second  degree,  257,  599. 
Inverse  of  a  transformation,  335. 
Involutory  transformations,  337. 
Isogonal   transformations,   336,   347, 

359. 

Kepter,  97. 

Latus  rectum,  94,  112,  135. 
Lemniscate,  201,  207,  215,  286. 
Limacon,  202,  215. 
Line,  equations  of, 

perpendicular    to    a  given    plane, 

483; 

parallel  to  two  planes,  484 ; 
parallel  to  a  given  line,  485 ; 
perpendicular  to  two  lines,  485. 
Line,    in    plane,    see   Contents,    Ch. 

II; 

equation    of   —   in    polar    coordi- 
nates, 196,  210; 
equation    of   —    in    normal    form, 

286.  287 ; 
equations    of    —    in    determinant 

form,  391,  393,  401 ; 
three    —    through    a    point,    53, 

189,  392 ; 

problems  in   three  —   through    a 

point,  54-60,  62,  189,  394 : 
single  equation  for  two  — ,    173, 

190,  191. 


Line,  in  space,  see  Contents,  Chs. 
XVIII,  XX,  XXI. 

Line-segments,  see  Contents,  Intro- 
duction, Chs.  I,  XVII. 

Linear  combination,  of  two  lines, 
165,  168,  169 ; 

—  of  two  curves,  167,  168,  170; 

—  of  two  planes,  498,  504,  506 ; 

—  of  two  surfaces,  520 ; 

—  of  three  planes,  521 ; 

—  of  sets  of  numbers,  400. 
Linear  equation,  in  x,  y,  31 ; 

—  in  x,  y,  z,  448. 

Linear  equation,  simultaneous,  see 
Contents,  Cb.  XVI,  §§8-10; 
also  401,  403,  464,  510. 

Loci  of  inequalities,  277,  279,  522. 

Loci    problems    in    plane,    see    Con- 
tents, Chs.  V,  XIII ; 
further  exercises  in  — ,   100,   123, 
152,  214,  328 ; 

—  in  polar  coordinates,  214. 

Loci   problems   in    space,   497,    546, 

582. 
Locus  of  an  equation,  19 ;    in  space, 

445; 

—  of  two  simultaneous  equations, 
471. 

Longitude,  584. 

Mid-point  of  a  line-segment,  16, 
415. 

Minimum  ellipse  of  an  unparted 
hyperboloid,  550. 

Minor  of  a  determinant,  367  ; 

use  of  —  to  evaluate  a  determi- 
nant, 367,  370,  380. 

Nappe  of  a  cone,  537. 
Nicomedes,  conchoid  of,  215. 
Nil-segment,  5. 

Normal  to  an  arbitrary  curve,  158 ; 
exercises  in   finding   equations   of 

— ,  160,  165. 
Normal,  to  a  parabola,  95 ; 

—  to  an  ellipse,  111,  112; 

—  to  a  hyperbola,  135. 
Normal,  to  a  plane,  447 ; 

direction    components    of    —    — , 
449; 

—  to   three   planes  parallel   to   a 
plane,  463 ; 


612 


INDEX 


—  to    four   planes   parallel    to    a 
plane,  508. 

Octants,  411. 

Optical  property  of  conies,  cf.  Focal 

property. 
Orbits  of  planets  and  comets,    120, 

211. 
Origin    of    rectangular    coordinates, 

7,410; 

—  of  polar  coordinates,  193. 
Orthogonality,  of  a  line  and  a  circle, 

76; 

—  of  two  circles,  76 ; 

—  of    two    confocal    conies,    95, 
146; 

—  of  a  sphere  and  a  plane,  545 ; 

—  of  a  sphere  and  a  line,  546 ; 

—  of  two  spheres,  546 ; 

—  of  confocal  quadrics,  591. 
Ovals  of  Cassini,  215. 

Parabola,  see  Contents,  Ch.  VI ; 

equations  of  —  not  in  normal 
form,  92,  97,  140,  230 ; 

equations  of  —  in  polar  coordi- 
nates, 203,  210,  211; 

equations  of  —  through  four 
points,  397,  399 ; 

eccentricity  of  — ,  140 ; 

—  as  limit  of  ellipses,  118 ; 

—  as  limit  of  hyperbolas,  139 ; 

—  as  locus  of  equation  of  the  sec- 
ond degree,  239,  254 ; 

confocal  — ,  95,  146. 
Paraboloid,  elliptic,  553 ; 

sections  of  — ,  553,  555,  563,  566 ; 

similar  — ,  581. 
Paraboloid,  hyperbolic,  553 ; 

sections  of  — ,  554,  555,  564,  580 ; 

rulings  of  — ,  559-561,  568,  582, 
583; 

directrix  planes  of  — ,  559 ; 

parametric  representation  of  — , 
560; 

similar  — ,  581. 

Paraboloid  of  revolution,  543,  553. 
Parabola,  semi-cubical,  231,  285. 
Parametric     representation,     of     an 
ellipse,  119,  306; 

—  of  a  hyperbola,  143,  308 ; 

—  of  a  line  in  space,  490 ; 


—  of  a  helix,  491 ; 

—  of  a  twisted  cubic,  492 ; 

—  of    an    unparted    hyperboloid, 
559; 

—  of    a    hyperbolic    paraboloid, 
560. 

Pascal,  limacon  of,  215. 
Pencil,  of  lines,  169,  172,  188 ; 

—  of  curves,  170,  172. 

Plane,  see  Contents,  Chs.  XIX,  XX, 

XXI. 
Plane,  equation  of, 

parallel    or    perpendicular    to     a 

given  plane,  456,  466,  467 ; 
perpendicular    to     a    given    line, 

483; 

parallel  to  two  lines,  484,  496; 
through  a  line  and  a  point,  499, 

504; 
through  a  line  parallel  to  a  second 

line,  513 ; 
determined  by  two  non-skew  lines, 

513. 
Planes,  three  through  a  line,  etc.,  cf. 

Condition,  in  space,  that. 
Pliicker,  abridged  notation  of,  171. 
Points,    three    on    a    line,    etc.,    cf. 

Condition,  in  the  plane,  that. 
Polar     coordinates,     see     Contents, 

Ch.  X; 
rotation  of  prime  direction  of  — , 

212; 
pole  of  —  in  an  arbitrary  point, 

213. 
Polar  lines  with  respect  to  a  quad- 

ric  surface,,  576. 
Pole  of  polar  coordinates,  193. 
Poles  and  polars,  with  respect  to  a 
conic,  see  Contents,  Ch.  XIV ; 

—  with   respect   to   a   degenerate 
conic,  402 ; 

—  with    respect    to    a    quadric 
surface,  575. 

Prime  direction,  193. 

Prime  meridian,  584. 

Principal   planes   of   a   quadric   sur- 
face, 549,  551,  555. 

Principal  sections  of  a  quadric  sur- 
face, 549,  551,  555. 

Product     of     two     transformations, 

338; 
exercises  in  — ,  341,  342,  358. 


INDEX 


613 


Projecting    planes    of    a    line,    501, 

503. 
Projection,  of  a  point  on  a  line,  5, 

405; 

—  of  a  directed  line-segment  on  a 
line,  5,  405,  408,  441 ; 

—  of  a  broken  line  on  a  line,  5, 
406; 

• —  of  a  directed  line-segment  on 
the  axes,  9,  413 ; 

—  of  a  point  on  a  plane,  406 ; 

—  of  a  line  on  a  plane,  406,  501- 
503; 

—  of   an   area   on   a   plane,    516, 
517; 

—  of  a  plane  curve  on  a  plane, 
534. 

Quadrants,  8. 

Quadratic  equation,  discriminant  of, 

175,  402 ; 

sum  of  roots  of  — ,  274. 
Quadric  cones,  539 ; 

sections  of  — ,  144,  539,  561-563, 

565. 

Quadric  cylinders,  532 ; 
sections  of  — ,  533 ; 
degenerate  — ,  602. 
Quadric  surfaces,  see  Contents,  Ch. 

XXIII ; 
confocal  — ,  590,  592 ; 

—  of  revolution,  540 ; 
similar  — ,  581. 

Radical  plane  of  two  spheres,  530. 
Radius  vector,  193,  584. 
Ray,   equation   of,   in   polar  coordi- 
nates, 196. 
Rectangular  hyperbola,  132 ; 

problems  concerning  — ,  151,  152, 

157; 
equation  of  —  in  polar  coordinates, 

207; 
equation  of  —  in    form  2xy  =  at, 

221; 
more    general    equations    of    — , 

239,  258,  259. 

Reflection  in  the  origin,  342. 
Reflections  in  the  axes,  336. 
Rigid  motion  of  the  plane,  358. 
Rotation    of    axes,    219,    239,    241 ; 
in  space,  593. 


Rotations  of  the  plane,  332. 

Rows  and  columns  of  a  determinant, 

363; 

interchanges  of  — ,  376,  380. 
Rulings,  of  a  cylinder,  532 ; 

—  of  a  cone,  536 ; 

—  of    an    unparted    hyperboloid, 
555-559,  568,  581-583 ; 

—  of    a    hyperbolic    paraboloid, 
559-561,  568,  582,  583. 

Sections,  of  quadric  cylinders,  533 ; 
1 —    of    quadric    cones,    144,    539, 
561-563,  565 ; 

—  of  a  quadric  surface,   defined, 
549; 

parallel  — ,  549,  561 ; 

circular ,  564 ; 

cf.  also  Ellipsoid,  etc. 
Shears,  simple,  351 ; 

factorization  of  — ,  355,  359. 
Shrinkings  of  the  plane,  334,  337. 
Similar  and  similarly  placed  conies, 

cf.  Conies. 
Similarity,  of  two  parabolas,  89 ; 

—  of  two  ellipses,  103,  121 ; 

—  of  two  hyperbolas,  128,  149 ; 

—  of  two  quadric  surfaces,  581. 
Similitude,   transformations  of,  334, 

358. 

Slope  angle,  13. 
Slope    of     a     curve,    definition     of, 

154; 
general  method  of  finding  — ,  154- 

158,  160-163; 

exercises  in  — ,  159,  164,  188. 
Slope,  of  a  line,  12 ; 

—  of  a  circle,  72 ; 

—  of  a  parabola,  94,  161 ; 

—  of  an  ellipse,  162 ; 

—  of  a  hyperbola,  163 ; 

—  of  a  general  conic,  188. 
Sphere,    see    Contents,    Ch.    XXII, 

§§1-5; 
null  — ,  524 ; 
further    problems    concerning    — , 

544-546 ; 

—  transformed     by     one-dimen- 
sional strains,  578,  579. 

Spherical  coordinates,  584,  590,  604. 
Spiral,  of  Archimedes,  202 ; 
— ,  hyperbolic,  202. 


614 


INDEX 


Square  array  of  a  determinant,  361, 

364. 

Square  root  sign,  11. 
Strains,  see  Contents,  Ch.  XV. 
Strains,    one-dimensional,    304,    307, 

309,   337;     in   space,   577. 
Strains,  homogeneous,  357. 
Strains,  simple,  357. 
Stretchings  of  the  plane,  334,  337. 
Supplemental  chords,  327. 
Surface,  definition  of,  444  ; 

symmetry  of  — ,  468. 
Surfaces  of  revolution,  540. 
Symmetry,  in  a  line,  83 ; 

—  in  a  point,  84 ; 

—  of  curves,  84-87 ;    in  polar  co- 
ordinates, 201 ; 

algebraic  — ,  267 ; 

—  of  surfaces,  468 ; 

—  of  quadric  surfaces,   600,   605, 
606. 

Tangent    at    a    given    point,    to    a 
circle,  69 ; 

—  to  a  parabola,  93,  164 ; 

—  to  an  ellipse,  111,  163; 

—  to  a  hyperbola,  133,  164,  180; 
•   —  to  a  general  conic,  188 ; 

construction  of  — ,  324. 
Tangent  line  to   a  quadric   surface, 

567. 
Tangent  plane,  to  a  sphere,  527 ; 

—  to  a  quadric  surface,  568,  580. 
Tangent   to   a   conic,    with   a   given 

slope,  174,  179; 

—  from  an  external  point,  185 ; 
construction  of  the  latter,  325 ; 
condition  that  a  line  be  — ,   183, 

192, 403 ; 

common  — ,  185,  191. 
Tangent     to     an     arbitrary     curve, 
definition  of,  154,  176; 


general    method    of    finding    slope 

of  — ,  154-158,  160-163 ; 
exercises  in   finding   equations   of 

— ,  158,  160,  165. 
Term  of  a  determinant,  366  ; 

determination  of  sign  of  — ,  365, 

366,  378. 
Tetrahedron,    center   of   gravity   of, 

419; 

volume  of  — ,  518. 
Torus,  544. 
Transformation  of  coordinates,  polar 

to  rectangular,  206,  214 ; 
— ,  spherical  to  rectangular,  585; 
— ,  cylindrical  to  rectangular,  587. 
Transformation    of    rectangular    co- 
ordinates,   see    Contents,    Chs. 
XI,  XXIV,  §§  5,  6. 
Transformation     to     parallel     axes, 

216,  235;   in  space,  592. 
Transformations  of  similitude,  334. 
Transformations    of    the    plane,    see 

Contents,  Ch.  XV. 
Translations,  330. 

Triangle,   area  of,   43 ;    in   determi- 
nant form,  401 ;   in  space,  517 ; 
medians  of  — ,  54,  56,  419 ; 
altitudes  of  — ,  58 ; 
perpendicular    bisectors    of    sides 

of—,  60; 

circle  circumscribing  — ,  74 ; 
circle  inscribed  in   — ,  283 ; 
nine-point  circle  of  — ,  78 ; 
bisectors  of  angles  of  — ,  283,  287. 
Triply   orthogonal    systems    of   sur- 
faces, 589-592. 

Umbilics,  580. 

Vertex  of  a  cone,  536. 
Vertex  of  a  paraboloid,  555. 
Vertices  of  a  conic,  88,  102,  125. 


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